标签：case Set No set each test input Odd

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a multiset (i. e. a set that can contain multiple equal integers) containing 2n2n integers. Determine if you can split it into exactly nn pairs (i. e. each element should be in exactly one pair) so that the sum of the two elements in each pair is odd (i. e. when divided by 22, the remainder is 11).

Input

The input consists of multiple test cases. The first line contains an integer tt (1≤t≤1001≤t≤100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer nn (1≤n≤1001≤n≤100).

The second line of each test case contains 2n2n integers a1,a2,…,a2na1,a2,…,a2n (0≤ai≤1000≤ai≤100) — the numbers in the set.

Output

For each test case, print "Yes" if it can be split into exactly nn pairs so that the sum of the two elements in each pair is odd, and "No" otherwise. You can print each letter in any case.

Example

input

Copy

5
2
2 3 4 5
3
2 3 4 5 5 5
1
2 4
1
2 3
4
1 5 3 2 6 7 3 4

output

Copy

Yes
No
No
Yes
No

Note

In the first test case, a possible way of splitting the set is (2,3)(2,3), (4,5)(4,5).

In the second, third and fifth test case, we can prove that there isn't any possible way.

In the fourth test case, a possible way of splitting the set is (2,3)(2,3).

解题说明：水题，统计数列中的奇偶数即可。

#include <stdio.h>
int main()
{
	int n, x, arr[200], a = 0, b = 0;
	scanf("%d", &n);
	for (int i = 0; i<n; i++, a = 0, b = 0)
	{
		scanf("%d", &x);
		for (int j = 0; j<2 * x; j++)
		{
			scanf("%d", &arr[j]);
			if (arr[j] % 2 == 0)
			{
				a++;
			}
			else
			{
				b++;
			}
		}
		if (a == b)
		{
			printf("Yes\n");
		}
		else
		{
			printf("No\n");
		}
	}
	return 0;
}

标签：case,Set,No,set,each,test,input,Odd
来源： https://blog.csdn.net/jj12345jj198999/article/details/118884573

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