#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <numeric>
#include <functional>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <sstream>
#include <string>
#include <cassert>
#define _repargs(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _repargs(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define mod 1000000007
#define inf 2000000007
#define mp make_pair
#define pb push_back
typedef long long ll;
using namespace std;
template <typename T>
inline void output(T a, int p = 0) {
    if(p) cout << fixed << setprecision(p)  << a << "\n";
    else cout << a << "\n";
}
// end of template

vector<ll> div(ll n){
    vector<ll> ret;
    for(ll d = 2; d * d <= n; d++) {
        if(n % d == 0) {
            ret.pb(d);
            if(d * d != n) ret.pb(n / d);
        }
    }
    ret.pb(n);
    sort(all(ret));
    return ret;
}

double EPS = 0.0000000000001;
vector<ll> D;

// nをD[m]以上のk個の約数で分解する方法
ll dec(ll n, ll m, int k){
    
    if(k == 0){
        if(n == 1) return 1;
        else return 0;
    }
    if(k == 1){
        if(n >= D[m]) return 1;
        else return 0;
    }
    if(pow(n, 1.0 / k) + EPS < D[m]) return 0;
    
    ll ret = 0;
    
    rep(d, m, D.size()){
        if(n < D[d] * D[d]) break;
        if(n % D[d] == 0){
            ll nn = n;
            int cnt = 1;
            while(nn % D[d] == 0 && k >= cnt){
                nn /= D[d];
                ret += dec(nn, d + 1, k - cnt);
//                cout << nn << ", " << D[d + 1] << ", " << k - cnt << ": " << dec(nn, d + 1, k - cnt) << endl;
                cnt++;
            }
        }
    }
//    cout << n << ", " << D[m] << ", " << k << ": " << ret << endl;
    
    return ret;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    // source code
    
    int N;
    ll X;
    cin >> N >> X;
    D = div(X + 1);
    ll M = dec(X + 1, 0, N);
    assert(M <= 200000);
    output(M);
    
    return 0;
}