A. Fancy Fence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.

He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.

Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal toa?

Input

The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.

Output

For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.

Sample test(s)

input

3
30
60
90

output

NO
YES
YES

Note

In the first test case, it is impossible to build the fence, since there is no regular polygon with angle .

In the second test case, the fence is a regular triangle, and in the last test case — a square.

解题说明：此题就是判断正多边形的角度取值问题，设边数为n，则内角和为180*n-360，问题转换为180-360/n=a是否有解，其中360能被n整除，循环遍历即可

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

int main()
{
	int t,n;
	int i,a;
	int j;
	int temp;
	scanf("%d",&t);
	for(i=0;i<t;i++)
	{
		scanf("%d",&a);
		for(j=3;j<=360;j++)
		{
			if(360%j==0)
			{
				temp=360/j;
				if(180-temp==a)
				{
					printf("YES\n");
					break;
				}
			}
		}
		if(j==361)
		{
			printf("NO\n");
		}
	}
    return 0;
}