Johnny has created a table which encodes the results of some operation -- a function of two arguments. But instead of a boring multiplication table of the sort you learn by heart at prep-school, he has created a GCD (greatest common divisor) table! So he now has a table (of height a and width b), indexed from (1,1) to (a,b), and with the value of field (i,j) equal to gcd(i,j). He wants to know how many times he has used prime numbers when writing the table.

Input

First, t ≤ 10, the number of test cases. Each test case consists of two integers, 1 ≤ a,b < 10⁷.

Output

For each test case write one number - the number of prime numbers Johnny wrote in that test case.

Example

Input:

2
10 10
100 100

Output:

30
2791

一样的题，仅仅只是 GCD(x,y) = 素数 .  1<=x<=a ; 1<=y<=b;

链接：http://www.spoj.com/problems/PGCD/ 

转载请注明出处：寻找&星空の孩子

具体解释：http://download.csdn.net/detail/u010579068/9034969

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int maxn=1e7+5;
typedef long long LL;
LL pri[maxn],pnum;
LL mu[maxn];
LL g[maxn];
LL sum[maxn];
bool vis[maxn];

void mobius(int N)
{
    LL i,j;
    pnum=0;
    memset(vis,false,sizeof(vis));
    vis[1]=true;
    mu[1]=1;
    for(i=2; i<=N; i++)
    {
        if(!vis[i])//pri
        {
            pri[pnum++]=i;
            mu[i]=-1;
            g[i]=1;
        }
        for(j=0; j<pnum && i*pri[j]<=N ; j++)
        {
            vis[i*pri[j]]=true;
            if(i%pri[j])
            {
                mu[i*pri[j]]=-mu[i];
                g[i*pri[j]]=mu[i]-g[i];
            }
            else
            {
                mu[i*pri[j]]=0;
                g[i*pri[j]]=mu[i];
                break;//think...
            }
        }
    }
    sum[0]=0;
    for(i=1; i<=N; i++)
    {
        sum[i]=sum[i-1]+g[i];
    }
}
int main()
{
    mobius(10000000);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        LL n,m;
        scanf("%lld%lld",&n,&m);
        if(n>m) swap(n,m);
        LL t,last,ans=0;
        for(t=1;t<=n;t=last+1)
        {
            last = min(n/(n/t),m/(m/t));
            ans += (n/t)*(m/t)*(sum[last]-sum[t-1]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}