1 /***
2 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
3 **/
4 #include <iostream>
5 #include <cstdio>
6 #include <algorithm>
7
8 using namespace std;
9 const int maxn = 50010;
10 int isprime[maxn],prime[maxn],mu[maxn],sum[maxn];
11
12 int t,a,b,c,d,k;
13 int cnt;
14 void mobius(int n){
15 int i,j;
16 cnt =0;
17 mu[1] =1;
18 for(i=2;i<=n;i++){
19 if(!isprime[i]){
20 prime[cnt++] = i;
21 mu[i] = -1;
22 }
23 for(j=0;j<cnt&&i*prime[j]<=n;j++){
24 isprime[i*prime[j]] = 1;
25 if(i%prime[j])
26 mu[i*prime[j] ] = -mu[i];
27 else{
28 mu[i*prime[j]] = 0;
29 break;
30 }
31 }
32 }
33 }
34
35 long long solve(int n,int m){
36 int i,la;
37 long long ret =0;
38 if(n>m)
39 swap(n,m);
40 for(i=1,la=0;i<=n;i=la+1){
41 la = min(n/(n/i),m/(m/i));
42 ret += (long long )(sum[la]-sum[i-1])*(n/i)*(m/i);
43 }
44 return ret;
45 }
46
47 int main(){
48 int i,j;
49 mobius(50000);
50 for(i=1;i<50000;i++)
51 sum[i] = sum[i-1]+mu[i];
52 scanf("%d",&t);
53 while(t--){
54 scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
55 long long ans;
56 ans = solve(b/k,d/k)-solve((a-1)/k,d/k)
57 -solve((c-1)/k,b/k)+solve((a-1)/k,(c-1)/k);
58 printf("%lld\n",ans);
59 }
60 return 0;
61 }
