Codeforces 900D (Codeforces Round #450 Div. 2) Unusual Sequences 容斥定理

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.

显然y必须是x的倍数，否则答案为0.

要想序列的gcd是x，则我们可以求出gcd是x的整数倍的序列个数，再利用容斥定理把它们减掉。

要求gcd为n*x的序列个数，我们可以把y分成k个n*x,再利用隔板法把它们隔开。这样，一共有2^(k-1)种序列。

若直接枚举n*x，个数太多。注意到n*x必须是y的因子，于是枚举y的因子缩小范围。

求出所有可能的因子之后，利用容斥定理从大到小推即可。

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