HYSBZ 2301

/***
对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数
**/
#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;
const int maxn = 50010;
int isprime[maxn],prime[maxn],mu[maxn],sum[maxn];

int t,a,b,c,d,k;
int cnt;
void mobius(int n){
    int i,j;
    cnt =0;
    mu[1] =1;
    for(i=2;i<=n;i++){
        if(!isprime[i]){
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(j=0;j<cnt&&i*prime[j]<=n;j++){
            isprime[i*prime[j]] = 1;
            if(i%prime[j])
                mu[i*prime[j] ] = -mu[i];
            else{
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}

long long solve(int n,int m){
    int i,la;
    long long ret =0;
    if(n>m)
        swap(n,m);
    for(i=1,la=0;i<=n;i=la+1){
        la = min(n/(n/i),m/(m/i));
        ret += (long long )(sum[la]-sum[i-1])*(n/i)*(m/i);
    }
    return ret;
}

int main(){
    int i,j;
    mobius(50000);
    for(i=1;i<50000;i++)
        sum[i] = sum[i-1]+mu[i];
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        long long ans;
        ans = solve(b/k,d/k)-solve((a-1)/k,d/k)
            -solve((c-1)/k,b/k)+solve((a-1)/k,(c-1)/k);
        printf("%lld
",ans);
    }
    return 0;
}