ecnu 3025

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

int solve(long long a,long long b,long long c){
    double u=b*b-4*a*c; //实际是解一个二元一次方程并求得大于等于2的整数解的个数(非0即1)
    if(u<0)return 0;
    long long v=(-b+sqrt(u))/2.0;
    if(v<=1)return 0;
    if(a*v*v+b*v+c==0)return 1;
    return 0;
}

int main(){
    long long t,m;
    scanf("%lld",&t);
    for(long long i=0;i<t;i++){
        scanf("%lld",&m);
        int cnt=0;
        for(long long j=1;j<=m/2;j++){
            cnt+=solve(1,2*j-1,-2*m);
        }
        printf("case #%lld:\n%lld\n",i,cnt);
    }
}