CodeForces - 808D Array Division

D. Array Division

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

3
1 3 2

output

Copy

YES

input

Copy

5
1 2 3 4 5

output

Copy

NO

input

Copy

5
2 2 3 4 5

output

Copy

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.

利用前缀和+二分查找来做

只要前半部分的和减一个数等于和的一半，或者后半部分的和加前半部分里的一个数等于和的一半，就证明可以

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
using namespace std;
typedef long long ll;

int a[100010];
ll sum[100010];

int main()
{
    int i,n;
    while(cin >> n){
        memset(sum,0,sizeof(sum));
        for(i = 1;i <= n;i++){
            cin >> a[i];
            sum[i] = sum[i-1] + a[i];
        }
        if(sum[n] % 2 != 0){//如果和是奇数，不可能分成两部分
            cout << "NO" << endl;
            continue;
        }
        ll m = sum[n]/2;
        int flag = 0;
        for(i = 1;i <= n;i++){//查sum而不是a，因为a无序，sum有序，而且sum要查两遍，从前往后和从后往前
            if(sum[lower_bound(sum+i,sum+n+1,m+a[i]) - sum] == m+a[i]){
                flag = 1;
                break;
            }
        }
        for(i = 1;i <= n;i++){
            if(sum[lower_bound(sum,sum+i-1,m-a[i]) - sum] == m-a[i]){//范围 错了好多次  重点
                flag = 1;
                break;
            }
        }
            if(flag) cout << "YES" << endl;
            else cout << "NO" << endl;
        }
    return 0;
}