C. Polygon for the Angle Educational Codeforces Round 57 (Rated for Div. 2)

C. Polygon for the Angle

题意：给你一个角度，让你求出存在此角度的最小正多边形，不存在输出-1；

题解:暴力，看代码注释

代码

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
#define int double
signed main(){
    int t;
    cin>>t;
    while(t--){
        int x;
        cin>>x;
        bool ok=0;
        for(int i=3;i<=998244353;i++){//暴力枚举每个正多边形
            int mx=(i-2)*180/i;/*最大内角*/
            int mn=(90-mx/2);/*最小内角*/
            //cout<<mx<<" "<<mn<<endl;
            for(int j=mn;j<=mx;j+=mn){/*依次枚举每一个可能的角,存在的角是等差数列，差是最小内角*/
                if(j==x){
                    ok=1;
                    break;
                }
            }
            if(ok){
                cout<<i<<endl;
                break;
            }
        }
        if(!ok){
            cout<<-1<<endl;
        }

    }
    return 0;
}

You are given an angle angang.

The Jury asks You to find such regular nn-gon (regular polygon with nn vertices) that it has three vertices aa, bb and cc (they can be non-consecutive) with ∠abc=ang∠abc=ang or report that there is no such nn-gon.

If there are several answers, print the minimal one. It is guarantied that if answer exists then it doesn't exceed 998244353998244353.

Input

The first line contains single integer TT (1≤T≤1801≤T≤180) — the number of queries.

Each of the next TT lines contains one integer angang (1≤ang<1801≤ang<180) — the angle measured in degrees.

Output

For each query print single integer nn (3≤n≤9982443533≤n≤998244353) — minimal possible number of vertices in the regular nn-gon or −1−1 if there is no such nn.

Example

input

Copy

4
54
50
2
178

output

Copy

10
18
90
180

Note

The answer for the first query is on the picture above.

The answer for the second query is reached on a regular 1818-gon. For example, ∠v2v1v6=50∘∠v2v1v6=50∘.

The example angle for the third query is ∠v11v10v12=2∘∠v11v10v12=2∘.

In the fourth query, minimal possible nn is 180180 (not 9090).