Codeforces Round #577 (Div. 2) B - Zero Array

Codeforces Round #577 (Div. 2)

 

B - Zero Array

You are given an array a1,a2,…,an.

In one operation you can choose two elements ai and aj (i≠j) and decrease each of them by one.

You need to check whether it is possible to make all the elements equal to zero or not.

Input

The first line contains a single integer n (2≤n≤105) — 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 it is possible to make all elements zero, otherwise print "NO".

Examples

input

4

1 1 2 2

output

YES

input

6

1 2 3 4 5 6

output

NO

Note

In the first example, you can make all elements equal to zero in 3 operations:

• Decrease a1 and a2,
• Decrease a3 and a4,
• Decrease a3 and a4

In the second example, one can show that it is impossible to make all elements equal to zero.

 

题意：题目意思是给你n个数，你可以进行任意次操作：对序列里的两个数都减一，

问你最后能不能把序列n个数都变成0。

思路：显然如果n个数的总和是奇数，肯定不能把n个数都变0。

还有就是序列中最大的数比总和的一半还大，即最大值的两倍大于总和，

这也是不行的，因为就算你拿另外n-1个数对最大值去操作，最后这个数还是不会变0，

然后其他情况都是可以的。

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<list>
//#include<unordered_map>
#include<stack>
using namespace std;
#define ll long long 
const int mod=1e9+7;
const int inf=1e9+7;
 
const int maxn=1e5+10;
 
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    
    int n;
    cin>>n;
    ll int num;
    ll int maxx=-1,sum;
    for(int i=0;i<n;i++)
    {
        cin>>num;
        sum+=num;
        
        if(num>maxx)
            maxx=num;
    }
    
    if(sum&1||maxx*2>sum)
        return cout<<"NO"<<endl,0;
    else
        return cout<<"YES"<<endl,0;
    
    return 0;
}