Description

 Today we have a number sequence A includes n elements. 
 Nero thinks a number sequence A is good only if the sum of its elements with odd index equals to the sum of its elements with even index and this sequence is not a palindrome. 
 Palindrome means no matter we read the sequence from head to tail or from tail to head,we get the same sequence. 
 Two sequence A and B are consider different if the length of A is different from the length of B or there exists an index i that $A_{i}\neq B_{i}$. 
 Now,give you the sequence A,check out it’s good or not.

Input

  The first line contains a single integer T,indicating the number of test cases. 
  Each test case begins with a line contains an integer n,the length of sequence A. 
 The next line follows n integers $A_{1},A_{2}, \ldots, A_{n}$. 

  [Technical Specification] 
  1 <= T <= 100 
  1 <= n <= 1000 
  0 <= $A_{i}$ <= 1000000

Output

  For each case output one line,if the sequence is good ,output "Yes",otherwise output "No".

Sample Input
3
7
1 2 3 4 5 6 7
7
1 2 3 5 4 7 6
6
1 2 3 3 2 1

Sample Output
No
Yes
No 

///本题就是让我们，统计给定序列奇数项的和是否为整个序列和的一半，并且这个序列为回文串，就输出Yes,否则为No，模拟就好。

//@zhangxiaoyu
//2015/8/12
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

int a[1010];
int main()
{
    int tt,n;
    bool flag;
    scanf("%d",&tt);
    while(tt--)
    {
        int sum1=0,sum2=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum1+=a[i];
        }
        for(int i=0;i<n;i+=2)
            sum2+=a[i];
        flag=true;
        for(int i=0,j=n-1;i<=j;i++,j--)
        {
            if(a[i]!=a[j])
            {
                flag=false;
                break;
            }
            continue;
        }
        if(sum2*2==sum1&&(flag==false))
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}