1 /**************************************************************
 2     Problem: 2820
 3     User: Tunix
 4     Language: C++
 5     Result: Accepted
 6     Time:4368 ms
 7     Memory:167304 kb
 8 ****************************************************************/
 9  
10 //BZOJ 2820
11 #include<cstdio>
12 #include<cstdlib>
13 #include<cstring>
14 #include<iostream>
15 #include<algorithm>
16 #define rep(i,n) for(int i=0;i<n;++i)
17 #define F(i,j,n) for(int i=j;i<=n;++i)
18 #define D(i,j,n) for(int i=j;i>=n;--i)
19 using namespace std;
20  
21 int getint(){
22     int v=0,sign=1; char ch=getchar();
23     while(ch<‘0‘||ch>‘9‘) {if (ch==‘-‘) sign=-1; ch=getchar();}
24     while(ch>=‘0‘&&ch<=‘9‘) {v=v*10+ch-‘0‘; ch=getchar();}
25     return v*=sign;
26 }
27 /*******************tamplate********************/
28 const int N=10001000;
29 typedef long long LL;
30 int mu[N],prime[N],g[N],sum[N];
31 bool check[N];
32  
33 void getmu(){
34     int tot=0;
35     mu[1]=1;
36     for(int i=2;i<N;i++){
37         if (!check[i]){
38             prime[tot++]=i;
39             mu[i]=-1;
40         }
41         rep(j,tot){
42             if(i*prime[j]>=N)break;
43             check[i*prime[j]]=1;
44             if(i%prime[j]==0){
45                 mu[i*prime[j]]=0;
46                 break;
47             }
48             else
49                 mu[i*prime[j]]=-mu[i];
50         }
51     }
52     rep(j,tot)
53         for(int i=prime[j];i<N;i+=prime[j])
54             sum[i]+=mu[i/prime[j]];//对于每个质数，枚举它的倍数 
55     for(int i=1;i<N;i++) sum[i]+=sum[i-1];
56 }
57  
58 int main(){
59     getmu();
60     int T=getint();
61     LL ans;
62     while(T--){
63         int n=getint(),m=getint();
64         if(n>m) swap(n,m);
65         ans=0;
66         for(int i=1,last;i<=n;i=last+1){
67             last=min(n/(n/i),m/(m/i));
68             ans+=(LL)(n/i)*(m/i)*(sum[last]-sum[i-1]);
69         }
70         printf("%lld\n",ans);
71     }
72     return 0;
73 }
74