#include<bits/stdc++.h>
using namespace std;

#define FOR(i,s,t) for(int i=(s),_t=(t); i<=_t; ++i)
#define DOR(i,s,t) for(int i=(s),_t=(t); i>=_t; --i)
#define EOR(i,x) for(int i=Head[x]; ~i; i=Nxt[i])

typedef double db;
typedef long long ll;

const int N=85;

int ans[]={0,1,2,2,4,2,4,4,8,2,4,6,8,2,8,6,16,2,4,6,8,4,12,6,16,4,4,4,16,2,12,10,32,4,4,8,8,2,12,6,16,2,8,6,24,6,12,8,32,6,8,6,8,2,8,10,32,4,4,6,24,2,20,6,64,6,8,8,8,4,16,6,16,2,4,8,24,14,12,6,32};
int a[N];
int n;
int dfs(int x,int d) {
    if(d==n) return 1;
    int y=(x+d)%n;
    int t=0;
    if(!a[y]) {
        a[y]=1;
        t+=dfs(y,d+1);
        a[y]=0;
    }
    y=(x-d+n)%n;
    if(!a[y]) {
        a[y]=1;
        t+=dfs(y,d+1);
        a[y]=0;
    }
    return t;
}
void solve() {
    scanf("%d",&n);
//    a[0]=1;
//    dfs(0,1);
    printf("%d\n",ans[n]);
}
int main() {
//    freopen("1.out","w",stdout);
//    FOR(i,1,80) {
//        n=i;
//        a[0]=1;
//        dfs(0,1);
//        printf("%d,",dfs(0,1)); 
//    }
    int T;
    scanf("%d",&T);
    while(T--) solve();
    return 0;
}