codeforces900D Unusual Sequences

D. Unusual Sequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Count the number of distinct sequences a1, a2, ..., an (1 ≤ ai) consisting of positive integers such that gcd(a1, a2, ..., an) = 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

Copy

3 9

output

input

Copy

5 8

output

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：问能构造出多少个序列满足，序列中所有元素的gcd为x，序列中所有元素的和为y

题解：首先序列中每个数字都可以用x*k表示，所以y是x的倍数，先判定。

之后就是放球问题，j个球放到i个箱子里允许为空。但是我们还要减去gcd是2x，3x...的方案数，这些数字正好可以对x分解质因数，然后逐个求解并减去。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[10005],dp[10005],n,m,tot;
ll pow(ll mi)
{ll res=1;for(ll a=2;mi;mi>>=1){if(mi&1) res=res*a%1000000007;a=a*a%1000000007;}return res;
}
int main()
{scanf("%I64d%I64d",&n,&m);if(m%n) return puts("0"),0;for(ll i=1;i<=sqrt(m);i++)if(m%i==0){if(i%n==0) a[++tot]=i;if(i*i!=m&&m/i%n==0) a[++tot]=m/i;}sort(a+1,a+tot+1);for(int i=tot;i;i--){dp[i]=pow(m/a[i]-1);for(int j=i+1;j<=tot;j++)if(a[j]%a[i]==0)dp[i]=(dp[i]-dp[j]+1000000007)%1000000007;}printf("%I64d\n",dp[1]);
}

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