題干：

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 to?a?

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.

Examples

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.

?

題目大意：

一條線，每次只能逆時針轉a度，問你是否可以恰好轉回來構成一個多邊形。（可以用來訓練一下讀題？）

解題報告：

? ?用180-a度，轉化成內角的度數，判斷360可否正好整除這個度數就可以了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5;int main() {int t,x;cin>>t;while(t--) {scanf("%d",&x);x=180-x;if(360%x == 0) puts("YES");else puts("NO"); }return 0 ;}

?

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