#include<bits/stdc++.h>
using namespace std;
#define ALL(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl;
#define INF 1000000000
#define mod 1000000007
using ll=long long;
const ll LINF=1001002003004005006ll;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>bool chmin(T &a,const T &b){if(b<a){a=b;return true;}return false;}

const int MAX=1500000;
long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % mod;
        //???
        inv[i] = mod - inv[mod%i] * (mod / i) % mod;
        finv[i] = finv[i - 1] * inv[i] % mod;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
}


//n!
ll fact_mod(ll n) {
    ll ret=1; 
    for(ll i=2;i<=n;i++) ret=ret*(i%mod)%mod;
    return ret;
}
 
// 繰り返し二乗法 
ll pow_mod(ll x, ll n){
    if(n==0) return 1;
    ll ret=pow_mod((x*x) % mod, n/2);
    if(n&1) ret=(ret*x)%mod;
    return ret;
}
 
//nCr O(r) nがでかくても安心
ll combination_mod(ll n, ll r) {
    if(r>n-r) r=n-r;
    if(r==0) return 1;
    ll a=1;
    //a=n!/(n-r)!=n~n-r+1までの総積->O(r)
    for(ll i=0;i<r;i++) a=a*((n-i)%mod)%mod;
    //b=inv(r!)
    ll b=pow_mod(fact_mod(r), mod-2);
    return (a%mod)*(b%mod)%mod;
}
 
ll inv_mod(ll n){
    // フェルマーの小定理
    return pow_mod(n,mod-2);
}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    COMinit();

    ll n,k;cin>>n>>k;
    vector<ll> dp(n+1,0);
    for(int i=1;i<n;i++)if(n%i==0){
        int t=n/i;//t個に分割
        if(k%t==0) dp[i]=COM(i,k/t);
    }

    ll ans=0;
    for(int i=1;i<n;i++)if(n%i==0){
        int t=n/i;
        if(k%t==0){
            ans+=dp[i];
            ans%=mod;
            for(int j=i+i;j<n;j+=i){
                dp[j]+=mod-dp[i];
                dp[j]%=mod;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}