超级大水题(还是自己过不了的水题)

题目链接:https://vjudge.net/contest/231314#problem/D

Peter Parker wants to play a game with Dr. Octopus. The game is about cycles. Cycle is a sequence of vertices, such that first one is connected with the second, second is connected with third and so on, while the last one is connected with the first one again. Cycle may consist of a single isolated vertex.

Initially there are k cycles, i-th of them consisting of exactly v_i vertices. Players play alternatively. Peter goes first. On each turn a player must choose a cycle with at least 2 vertices (for example, x vertices) among all available cycles and replace it by two cycles with p and x - p vertices where 1 ≤ p < x is chosen by the player. The player who cannot make a move loses the game (and his life!).

Peter wants to test some configurations of initial cycle sets before he actually plays with Dr. Octopus. Initially he has an empty set. In the i-th test he adds a cycle with a_i vertices to the set (this is actually a multiset because it can contain two or more identical cycles). After each test, Peter wants to know that if the players begin the game with the current set of cycles, who wins?

Peter is pretty good at math, but now he asks you to help.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of tests Peter is about to make.

The second line contains n space separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹), i-th of them stands for the number of vertices in the cycle added before the i-th test.

Output

Print the result of all tests in order they are performed. Print 1 if the player who moves first wins or 2 otherwise.

Examples

Input

3
1 2 3

Output

2
1
1

Input

5
1 1 5 1 1

Output

2
2
2
2
2

Note

In the first sample test:

In Peter's first test, there's only one cycle with 1 vertex. First player cannot make a move and loses.

In his second test, there's one cycle with 1 vertex and one with 2. No one can make a move on the cycle with 1 vertex. First player can replace the second cycle with two cycles of 1 vertex and second player can't make any move and loses.

In his third test, cycles have 1, 2 and 3 vertices. Like last test, no one can make a move on the first cycle. First player can replace the third cycle with one cycle with size 1 and one with size 2. Now cycles have 1, 1, 2, 2 vertices. Second player's only move is to replace a cycle of size 2 with 2 cycles of size 1. And cycles are 1, 1, 1, 1, 2. First player replaces the last cycle with 2 cycles with size 1 and wins.

In the second sample test:

Having cycles of size 1 is like not having them (because no one can make a move on them).

In Peter's third test: There a cycle of size 5 (others don't matter). First player has two options: replace it with cycles of sizes 1 and 4 or 2 and 3.

• If he replaces it with cycles of sizes 1 and 4: Only second cycle matters. Second player will replace it with 2 cycles of sizes 2. First player's only option to replace one of them with two cycles of size 1. Second player does the same thing with the other cycle. First player can't make any move and loses.
• If he replaces it with cycles of sizes 2 and 3: Second player will replace the cycle of size 3 with two of sizes 1 and 2. Now only cycles with more than one vertex are two cycles of size 2. As shown in previous case, with 2 cycles of size 2 second player wins.

So, either way first player loses.

题目大意:n代表有n个测试样例,下面n组数，注意后面的数的结果是在前面数的结果之上出来的，大概意思就是给你一组数，假设一个数是

x,你可以用两个数来替换它，p,x-p,把所有数换成1的是谁，先走的赢输出1，后走的赢输出2。

个人思路:以为是到很难的博弈题，然后在草稿本上推公式，应该是推出来了，但是超时了，然后优化，优化了之后变成runtime了，数组范围

不够，但是又不能开的要的那么大，所以过不了，只能搜题解了。 题解很简单，就是一个数要变成1的话要n-1次，所以只要统计有多少次就行

了，是不是特别水的题，，，，

看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
const int maxn=1e5+10;
const ll maxa=1e8;
#define INF 0x3f3f3f
//#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int main()
{
    int n;
    ll sum=0;
    ll a[maxn];
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i];
    for(int i=0;i<n;i++)
    {
        sum+=a[i]-1;
        if(sum%2)
            cout<<"1"<<endl;
        else
            cout<<"2"<<endl;
    }
    return 0;
}

下面这道是我见过最水的题，没有之一

题目链接：http://codeforces.com/problemset/problem/710/B

You are given n points on a line with their coordinates x_i. Find the point x so the sum of distances to the given points is minimal.

Input

The first line contains integer n (1 ≤ n ≤ 3·10⁵) — the number of points on the line.

The second line contains n integers x_i ( - 10⁹ ≤ x_i ≤ 10⁹) — the coordinates of the given n points.

Output

Print the only integer x — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer.

Example

Input

4
1 2 3 4

Output

2
题目大意：输入n,代表 有n个坐标，下面给出n个坐标，求哪个点所有点到它的距离和最小
个人思路：以为给出的是各个点的大小，给出顺序是坐标，想的挺复杂，然后完全偏离方向，嗯。。。  很水

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
const int maxn=3e5+10;
const ll maxa=1e10;
#define INF 0x3f3f3f3f3f3f
int a[maxn],b[maxn];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    sort(a,a+n);
    if(n%2==0)
        cout<<a[n/2-1];
    else
        cout<<a[n/2];
    return 0;
}