CodeForces

Count the number of distinct sequences a₁, a₂, ..., a_n (1 ≤ a_i) consisting of positive integers such that gcd(a₁, a₂, ..., a_n) = x and . As this number could be large, print the answer modulo 10⁹ + 7.

gcd here means the greatest common divisor.

Input

The only line contains two positive integers x and y (1 ≤ x, y ≤ 10⁹).

Output

Print the number of such sequences modulo 10⁹ + 7.

Examples

Input

3 9

Output

3

Input

5 8

Output

0

Note

There are three suitable sequences in the first test: (3, 3, 3), (3, 6), (6, 3).

There are no suitable sequences in the second test.

题意：N个未知数，他们的GCD是X，和是Y，问方案数。

思路：显然我们枚举GCD=x，然后就是(Y/X)/x-1个隔板，2^隔板。前面加莫比乌斯系数mu[x/X]即可。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int Mod=1e9+7;
int p[20],cnt,ans,X,Y;
int qpow(int a,int x){
    int res=1; while(x){
        if(x&1) res=(ll)res*a%Mod;
        a=(ll)a*a%Mod; x>>=1;
    }return res;
}
void dfs(int pos,int s,int opt)
{
    if(pos>cnt) { ans=((ans+opt*qpow(2,Y/s-1))%Mod+Mod)%Mod; return ;}
    dfs(pos+1,s,opt); dfs(pos+1,s*p[pos],-opt);
}
int main()
{
    scanf("%d%d",&X,&Y);
    if(Y%X!=0) return puts("0"),0;
    Y/=X; int tp=Y;
    for(int i=2;i<=tp/i;i++){
        if(tp%i==0) {
            p[++cnt]=i;
            while(tp%i==0) tp/=i;
        }
    }
    if(tp>1) p[++cnt]=tp;
    dfs(1,1,1);
    printf("%d
",ans);
    return 0;
}