Educational Codeforces Round 21 (D. Array Division)

Vasya has an array a consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).

Inserting an element in the same position he was erased from is also considered moving.

Can Vasya divide the array after choosing the right element to move and its new position?

Input

The first line contains single integer n (1 ≤ n ≤ 100000) — the size of the array.

The second line contains n integers a1, a2... an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print YES if Vasya can divide the array after moving one element. Otherwise print NO.

Examples

input

3
1 3 2

output

YES

input

5
1 2 3 4 5

output

NO

input

5
2 2 3 4 5

output

YES

Note

In the first example Vasya can move the second element to the end of the array.

In the second example no move can make the division possible.

In the third example Vasya can move the fourth element by one position to the left.

题意：是否可以移动一个元素使得数组分为两个相等的部分(即存在前缀和等于后缀和)。

分为两种情况，1.移动某个元素到右边，2.移动某个元素到左边，3.不需要移动。

第三种情况比较简单就不说了，求出k=sum/2,最后希望的是使得1.前缀和移动一个元素到后面使得前缀和等于k 或者 2.后缀和移动一个元素到前面使得后缀和等于k。

即prefix[i]-num[j]=k(j<=i)或者suffix[i]-num[j]=k(j>=i)，即prefix[i]=k+num[j](j<=i)或者suffix[i]=k+num[j](j>=i)。

用map记录一下k+num[j]的位置j,在判断一下是否存在prefix[i]或者suffix[i]是否有效即可。

注意一下可能出现k+num[j1]=k+num[j2]的情况，所以判断前缀的时候map记录min(j1,j2),判断后缀的时候map记录max(j1,j2)。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define max(x,y) (x)>(y)?(x):(y)
#define min(x,y) (x)>(y)?(y):(x)
#define cls(name,x) memset(name,x,sizeof(name))
using namespace std;
const int inf=1<<28;
const int maxn=100010;
const int maxm=110;
const int mod=1e9+7;
const double pi=acos(-1.0);
int n;
ll num[maxn];
ll sum[maxn];
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        int flag=0;
        sum[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&num[i]);
            sum[i]=sum[i-1]+num[i];
        }
        if(sum[n]%2==1)
        {
            printf("NO
");
            continue;
        }
        ll k=sum[n]/2;
        for(int i=1;i<=n;i++)
            if(sum[i]==k)
            {flag=1;break;}
        map<ll,int> mapp;
        for(int i=1;i<=n;i++)
        {
            if(mapp[k+num[i]]!=0)
            mapp[k+num[i]]=min(i,mapp[k+num[i]]);
            else
                mapp[k+num[i]]=i;
        }
        for(int i=1;i<=n;i++)
        {
            if(mapp[sum[i]]!=0&&i>=mapp[sum[i]])
                flag=1;
        }
        for(int i=1;i<=n;i++)
        {
            mapp[k+num[i]]=max(i,mapp[k+num[i]]);
        }
        for(int i=n;i>=1;i--)
        {
            if(mapp[sum[n]-sum[i-1]]!=0&&i<=mapp[sum[n]-sum[i-1]])
                flag=1;
        }
        if(flag) printf("YES
");
        else printf("NO
");
    }
    return 0;
}