/* -*- coding: utf-8 -*-
 *
 * 890.cc:  No.890 移調の限られた旋法 - yukicoder
 */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

const int MAX_N = 1000000;
const int MAX_M = 8;
const int MOD = 1000000007;

/* typedef */

typedef long long ll;

/* global variables */

bool primes[MAX_N + 1];
int pnums[MAX_N], dps[MAX_M];
int fracs[MAX_N + 1], invfs[MAX_N + 1];

/* subroutines */

template <typename T>
T gcd(T m, T n) {  // m > 0, n > 0
  if (m < n) swap(m, n);
  while (n > 0) {
    T r = m % n;
    m = n;
    n = r;
  }
  return m;
}

int gen_primes(int maxp, int pnums[]) {
  memset(primes, true, sizeof(primes));
  primes[0] = primes[1] = false;

  int p, pn = 0;
  for (p = 2; p * p <= maxp; p++)
    if (primes[p]) {
      pnums[pn++] = p;
      for (int q = p * p; q <= maxp; q += p) primes[q] = false;
    }
  for (; p <= maxp; p++)
    if (primes[p]) pnums[pn++] = p;
  return pn;
}

int powmod(int a, int n) {  // a^n % MOD
  int pm = 1;
  while (n > 0) {
    if (n & 1) pm = (ll)pm * a % MOD;
    a = (ll)a * a % MOD;
    n >>= 1;
  }
  return pm;
}

inline int invf(int n) { return powmod(fracs[n], MOD - 2); }

inline int nck(int n, int k) {  // nCk % MOD
  return (ll)fracs[n] * invf(n - k) % MOD * invf(k) % MOD;
}

/* main */

int main() {
  int n, k;
  scanf("%d%d", &n, &k);

  int g = gcd(n, k);
  int pn = gen_primes(g, pnums);

  int m = 0;
  for (int i = 0; i < pn; i++)
    if (g % pnums[i] == 0) dps[m++] = pnums[i];
  //printf("m=%d\n", m);

  fracs[0] = 1;
  for (int i = 1; i <= n; i++) fracs[i] = (ll)fracs[i - 1] * i;

  int sum = 0, mbits = 1 << m;
  for (int bits = 1; bits < mbits; bits++) {
    int bn = 0, d = 1;
    for (int i = 0, bi = 1; i < m; i++, bi <<= 1)
      if (bits & bi) bn++, d *= dps[i];

    int c = nck(n / d, k / d);
    sum = (sum + ((bn & 1) ? c : MOD - c)) % MOD;
  }

  printf("%d\n", sum);
  return 0;
}