A. Array with Odd Sum_you are given an array aa consisting of nn integer

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array aa consisting of nn integers.

In one move, you can choose two indices 1≤i,j≤n1≤i,j≤n such that i≠ji≠j and set ai:=ajai:=aj. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose ii and jj and replace aiai with ajaj).

Your task is to say if it is possible to obtain an array with an odd (not divisible by 22) sum of elements.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤20001≤t≤2000) — the number of test cases.

The next 2t2t lines describe test cases. The first line of the test case contains one integer nn (1≤n≤20001≤n≤2000) — the number of elements in aa. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤20001≤ai≤2000), where aiai is the ii-th element of aa.

It is guaranteed that the sum of nn over all test cases does not exceed 20002000 (∑n≤2000∑n≤2000).

Output

For each test case, print the answer on it — "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.

Example

input

Copy

5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1

output

Copy

YES
NO
YES
NO
NO

解题说明：水题，判断数组中是否存在奇数即可，可以用奇数替换掉其中的偶数，这样能保证数组之和肯定为奇数，但要排除偶数个奇数的情况。

 
 
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
 
using namespace std;
 
int main() 
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int n, c = 0;
		scanf("%d", &n);
		int a[2002];
		for (int i = 0; i<n; i++)
		{
			scanf("%d", &a[i]);
			if (a[i] % 2 != 0)
			{
				c++;
			}
		}
		if ((c == n && n % 2 == 0) || c == 0)
		{
			printf("NO\n");
		}
		else
		{
			printf("YES\n");
		}
	}
	return 0;
}