Codeforces 900D-Unusual Sequences

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：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

解题思路：容斥，分别求出gcd为x，2*x……的数量，然后减去多余即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>#include <ctime>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const LL mod = 1e9 + 7;map<int, LL>mp;LL qpow(int k){LL ans = 1, x = 2;while (k){if (k & 1) (ans *= x) %= mod;(x *= x) %= mod;k >>= 1;}return ans;}LL solve(int x){if (x == 1) return 1;if (mp[x]) return mp[x];mp[x] = qpow(x - 1);for (int i = 2; i*i <= x; i++){if (x%i == 0){mp[x] = (mp[x] - solve(x / i) + mod) % mod;if (i != x / i) mp[x] = (mp[x] - solve(i) + mod) % mod;}}mp[x] = (mp[x] - solve(1) + mod) % mod;return mp[x];}int main(){int x, y;while (~scanf("%d%d", &x, &y)){if (y%x != 0) { printf("0\n"); continue; }mp.clear();printf("%lld\n", solve(y / x));}return 0;}