Manthan, Codefest 17

Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.

He has names of n people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.

Formally, for a name si in the i-th line, output "YES" (without quotes) if there exists an index j such that si = sj and j < i, otherwise, output "NO" (without quotes).

Input

First line of input contains an integer n (1 ≤ n ≤ 100) — the number of names in the list.

Next n lines each contain a string si, consisting of lowercase English letters. The length of each string is between 1 and 100.

Output

Output n lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.

You can print each letter in any case (upper or lower).

Examples

input

6
tom
lucius
ginny
harry
ginny
harry

output

NO
NO
NO
NO
YES
YES

input

3
a
a
a

output

NO
YES
YES

Note

In test case 1, for i = 5 there exists j = 3 such that si = sj and j < i, which means that answer for i = 5 is "YES".

直接枚举啦，就是判断出现过没

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin >> n;
    set<string> s;
    for (int i = 0; i < n; i++)
    {
        string c;
        cin >> c;
        if (s.count(c)) puts("YES");
        else puts("NO");
        s.insert(c);
    }
    return 0;
}

Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly x drops of the potion he made.

Value of x is calculated as maximum of p·ai + q·aj + r·ak for given p, q, r and array a1, a2, ... an such that 1 ≤ i ≤ j ≤ k ≤ n. Help Snape find the value of x. Do note that the value of x may be negative.

Input

First line of input contains 4 integers n, p, q, r ( - 109 ≤ p, q, r ≤ 109, 1 ≤ n ≤ 105).

Next line of input contains n space separated integers a1, a2, ... an ( - 109 ≤ ai ≤ 109).

Output

Output a single integer the maximum value of p·ai + q·aj + r·ak that can be obtained provided 1 ≤ i ≤ j ≤ k ≤ n.

Examples

input

5 1 2 3
1 2 3 4 5

output

input

5 1 2 -3
-1 -2 -3 -4 -5

output

Note

In the first sample case, we can take i = j = k = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.

In second sample case, selecting i = j = 1 and k = 5 gives the answer 12.

Create a dynamic programming table of size n·3. In this, dp[i][0] stores maximum of value p·ax for x between 1 and i. Similarly dp[i][1] stores the maximum value of p·ax + q·ay such that x ≤ y ≤ i and dp[i][2] stores maximum value of p·ax + q·ay + r·az for x ≤ y ≤ z ≤ i.

To calculate the dp:

dp[i][0] = max(dp[i - 1][0], p·ai)

dp[i][1] = max(dp[i - 1][1], dp[i][0] + q·ai)

dp[i][2] = max(dp[i - 1][2], dp[i][1] + r·ai)

The answer will be stored in dp[n][2]

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+5;
ll a[N];
ll ans[3];
int main()
{
    ll n,p,q,r;
    cin>>n>>p>>q>>r;
    for(int i=1; i<=n; i++)
        cin>>a[i];
    for(int i=0; i<3; i++)
        ans[i]=-9e18;
    for(int i=1; i<=n; i++)
    {
        ans[0]=max(ans[0],a[i]*p);
        ans[1]=max(ans[1],ans[0]+q*a[i]);
        ans[2]=max(ans[2],ans[1]+r*a[i]);
    }
    cout<<ans[2]<<endl;
    return 0;
}

Harry, Ron and Hermione have figured out that Helga Hufflepuff's cup is a horcrux. Through her encounter with Bellatrix Lestrange, Hermione came to know that the cup is present in Bellatrix's family vault in Gringott's Wizarding Bank.

The Wizarding bank is in the form of a tree with total n vaults where each vault has some type, denoted by a number between 1 to m. A tree is an undirected connected graph with no cycles.

The vaults with the highest security are of type k, and all vaults of type k have the highest security.

There can be at most x vaults of highest security.

Also, if a vault is of the highest security, its adjacent vaults are guaranteed to not be of the highest security and their type is guaranteed to be less than k.

Harry wants to consider every possibility so that he can easily find the best path to reach Bellatrix's vault. So, you have to tell him, given the tree structure of Gringotts, the number of possible ways of giving each vault a type such that the above conditions hold.

Input

The first line of input contains two space separated integers, n and m — the number of vaults and the number of different vault types possible. (1 ≤ n ≤ 105, 1 ≤ m ≤ 109).

Each of the next n - 1 lines contain two space separated integers ui and vi (1 ≤ ui, vi ≤ n) representing the i-th edge, which shows there is a path between the two vaults ui and vi. It is guaranteed that the given graph is a tree.

The last line of input contains two integers k and x (1 ≤ k ≤ m, 1 ≤ x ≤ 10), the type of the highest security vault and the maximum possible number of vaults of highest security.

Output

Output a single integer, the number of ways of giving each vault a type following the conditions modulo 109 + 7.

Examples

input

output

input

output

input

output

Note

In test case 1, we cannot have any vault of the highest security as its type is 1 implying that its adjacent vaults would have to have a vault type less than 1, which is not allowed. Thus, there is only one possible combination, in which all the vaults have type 2.

这个题我有点懵

树形dp，类似于染色吧

#include <bits/stdc++.h>
using namespace std;
const int  N=1e5+5;
const int MD=1e9+7;
typedef long long LL;
int n, m, k, x;
int f[N][11][3], g[11][3];
vector < int > V[N];
void DFS(int u, int fa)
{
    f[u][0][0] = k - 1;
    f[u][1][1] = 1;
    f[u][0][2] = m - k;
    for(auto &v: V[u])
    {
        if(v == fa) continue;
        DFS(v, u);
        memset(g, 0, sizeof(g));
        for(int i = x; i>=0; --i)
            for(int j=x-i; j>=0; --j)
            {
                g[i + j][0] = (g[i + j][0] + 1LL * f[u][i][0] * f[v][j][0]) % MD;
                g[i + j][0] = (g[i + j][0] + 1LL * f[u][i][0] * f[v][j][1]) % MD;
                g[i + j][0] = (g[i + j][0] + 1LL * f[u][i][0] * f[v][j][2]) % MD;
                g[i + j][1] = (g[i + j][1] + 1LL * f[u][i][1] * f[v][j][0]) % MD;
                g[i + j][2] = (g[i + j][2] + 1LL * f[u][i][2] * f[v][j][0]) % MD;
                g[i + j][2] = (g[i + j][2] + 1LL * f[u][i][2] * f[v][j][2]) % MD;
            }
        for(int i = 0; i <= x; ++i)
            for(int j = 0; j < 3; ++j)
                f[u][i][j] = g[i][j];
    }
}
int main()
{
    scanf("%d %d", &n, &m);
    for(int i = 1; i < n; ++i)
    {
        int u,v;
        scanf("%d %d", &u, &v);
        V[u].push_back(v);
        V[v].push_back(u);
    }
    scanf("%d %d", &k, &x);
    DFS(1, 0);
    int ans = 0;
    for(int i = 0; i <= x; ++i)
        for(int j = 0; j < 3; ++j)
            ans = (ans + f[1][i][j]) % MD;
    printf("%d
", ans);
    return 0;
}