C. Odd/Even Increments

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Given an array a=[a1,a2,…,an]a=[a1,a2,…,an] of nn positive integers, you can do operations of two types on it:

Add 11 to every element with an odd index. In other words change the array as follows: a1:=a1+1,a3:=a3+1,a5:=a5+1,…a1:=a1+1,a3:=a3+1,a5:=a5+1,….
Add 11 to every element with an even index. In other words change the array as follows: a2:=a2+1,a4:=a4+1,a6:=a6+1,…a2:=a2+1,a4:=a4+1,a6:=a6+1,….

Determine if after any number of operations it is possible to make the final array contain only even numbers or only odd numbers. In other words, determine if you can make all elements of the array have the same parity after any number of operations.

Note that you can do operations of both types any number of times (even none). Operations of different types can be performed a different number of times.

Input

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

The first line of each test case contains an integer nn (2≤n≤502≤n≤50) — the length of the array.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1031≤ai≤103) — the elements of the array.

Note that after the performed operations the elements in the array can become greater than 103103.

Output

Output tt lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if after any number of operations it is possible to make the final array contain only even numbers or only odd numbers, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Example

input

Copy

4
3
1 2 1
4
2 2 2 3
4
2 2 2 2
5
1000 1 1000 1 1000

output

Copy

YES
NO
YES
YES

Note

For the first test case, we can increment the elements with an even index, obtaining the array [1,3,1][1,3,1], which contains only odd numbers, so the answer is "YES".

For the second test case, we can show that after performing any number of operations we won't be able to make all elements have the same parity, so the answer is "NO".

For the third test case, all elements already have the same parity so the answer is "YES".

For the fourth test case, we can perform one operation and increase all elements at odd positions by 11, thus obtaining the array [1001,1,1001,1,1001][1001,1,1001,1,1001], and all elements become odd so the answer is "YES".

解题说明：此题找规律能发现必须要保证数列奇数位置奇偶状态一致，偶数位置也是同样的道理，只需要判断前两个，后面遍历即可。

#include<stdio.h>
int main()
{int t, n, a[52], b1, b2, i, flag;scanf("%d", &t);while (t--) {flag = 0;scanf("%d", &n);scanf("%d", &a[1]);scanf("%d", &a[2]);b1 = a[1] % 2;b2 = a[2] % 2;if (n>2) {for (i = 3; i <= n; i++){scanf("%d", &a[i]);if (i % 2 == 1) {if (a[i] % 2 != b1){flag = 1;}}else {if (a[i] % 2 != b2){flag = 1;}}}}if (flag == 0){printf("YES\n");}else{printf("NO\n");}}return 0;
}

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