codeforces 900D Unusual Sequences （数论）

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

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：给出 x,y 让求 有几种组合使得他们的gcd为 x并且和为y

分析：这题有点类似于隔板法 

那么知道满足的最小条件时 y%x==0

然后将 y分成 p=y/x分 每一份就为 x且 满足条件

这个时候这 p份可以排列组合得到的也必定满足 和为 y

那么只需要容斥去掉 gcd不为 x的就可以了

AC代码：

#include
using namespace std;
const int mod=1000000007;
long long f(long long a,long long b)
{
	long long c=1;
	while(b)
	{
		if(b&1)
		{
			c*=a;
			c%=mod;
		}
		b>>=1;
		a*=a;
		a%=mod;
	}
	return c; 
}
long long vis[1000000];
long long dp[1000000]={0};
int main()
{
	long long x,y;
	scanf("%lld%lld",&x,&y);
	int t=0;
	if(y%x)
	printf("0\n");
	else
	{
		long long q=y/x;
		for(long long i=1;i*i<=q;i++)
		{
			if(q%i==0)
			{
				vis[t++]=i;
				if(i*i!=q)
				{
					vis[t++]=q/i;
					//printf("%lld\n",q/i);
				}
			}
		}
		sort(vis,vis+t);
		for(int i=0;i=0;i--)
		{
			for(int j=0;j