// Original C++ solution (slower):
#include <bits/stdc++.h>

#define ADD(a, b) a = (a + ll(b)) % mod

#define MUL(a, b) a = (a * ll(b)) % mod

#define MAX(a, b) a = max(a, b)

#define MIN(a, b) a = min(a, b)

#define rep(i, a, b) for(int i = int(a); i < int(b); i++)

#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)

#define all(a) (a).begin(), (a).end()

#define sz(v) (int)(v).size()

#define pb push_back

#define sec second

#define fst first

#define debug(fmt, ...) Debug(__LINE__, ":", fmt, ##__VA_ARGS__)

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pi;

typedef pair<ll, ll> pl;

typedef pair<int, pi> ppi;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef vector<vl> mat;

typedef complex<double> comp;

void Debug() {cout << '\n'; }

template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){

	cout<<arg<<" ";Debug(rest...);}

template<class T>ostream& operator<<(ostream& out,const vector<T>& v) {

	out<<"[";if(!v.empty()){rep(i,0,sz(v)-1)out<<v[i]<<", ";out<<v.back();}out<<"]";return out;}

template<class S, class T>ostream& operator<<(ostream& out,const pair<S, T>& v){

	out<<"("<<v.first<<", "<<v.second<<")";return out;}

const int MAX_N = 400010;

const int MAX_V = 100010;

const double eps = 1e-6;

const ll mod = 1000000007;

const int inf = 1 << 29;

const ll linf = 1LL << 60;

const double PI = 3.14159265358979323846;

///////////////////////////////////////////////////////////////////////////////////////////////////



ll fac[MAX_N], inv[MAX_N], fiv[MAX_N]; //fiv:inv(fac(i))



ll mod_pow(ll a, ll n) {

	if(n == 0) return 1;

	ll res = mod_pow(a, n / 2);

	if(n % 2 == 0) return res * res % mod;

	else return a * res % mod * res % mod;

}



ll invf(ll a) {

	return mod_pow(a, mod - 2);

}



void C_init(int n) {

	fac[0] = fac[1] = 1; inv[1] = 1;

	fiv[0] = fiv[1] = 1;

	rep(i, 2, n + 1) {

		fac[i] = fac[i - 1] * i % mod;

		inv[i] = mod - inv[mod % i] * (mod / i) % mod;

		fiv[i] = fiv[i - 1] * inv[i] % mod;

	}

}



ll C(int a, int b) { //assume a >= b

	if(a < b || a < 0 || b < 0) return 0;

	return fac[a] * fiv[b] % mod * fiv[a - b] % mod;

}



int N;

ll M;



void solve() {

	cin >> N >> M;

	C_init(2 * N + 100000);

	vector<ll> vec;

	for(ll i = 2; i * i <= M; i++) {

		if(M % i == 0) {

			int cnt = 0;

			while(M % i == 0) {

				cnt++;

				M /= i;

			}

			vec.pb(cnt);

		}

	}

	if(M != 1) vec.pb(1);

	ll res = 1;

	rep(i, 0, sz(vec)) {

		// debug(N - 1 + vec[i], vec[i]);

		MUL(res, C(N - 1 + vec[i], vec[i]));

	}

	cout << res << "\n";

}



int main() {

#ifndef LOCAL

	ios::sync_with_stdio(false);

    cin.tie(0);

#endif

    cout << fixed;

	cout.precision(20);

	srand((unsigned int)time(NULL));

#ifdef LOCAL

	//freopen("in.txt", "wt", stdout); //for tester

    freopen("in.txt", "rt", stdin);

#endif	

	solve();

#ifdef LOCAL

    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

#endif

	return 0;

}




// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

#define ADD(a, b) a = (a + ll(b)) % mod

#define MUL(a, b) a = (a * ll(b)) % mod

#define MAX(a, b) a = max(a, b)

#define MIN(a, b) a = min(a, b)

#define rep(i, a, b) for(int i = int(a); i < int(b); i++)

#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)

#define all(a) (a).begin(), (a).end()

#define sz(v) (int)(v).size()

#define pb push_back

#define sec second

#define fst first

#define debug(fmt, ...) Debug(__LINE__, ":", fmt, ##__VA_ARGS__)

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pi;

typedef pair<ll, ll> pl;

typedef pair<int, pi> ppi;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef vector<vl> mat;

typedef complex<double> comp;

void Debug() {cout << '\n'; }

template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){

	cout<<arg<<" ";Debug(rest...);}

template<class T>ostream& operator<<(ostream& out,const vector<T>& v) {

	out<<"[";if(!v.empty()){rep(i,0,sz(v)-1)out<<v[i]<<", ";out<<v.back();}out<<"]";return out;}

template<class S, class T>ostream& operator<<(ostream& out,const pair<S, T>& v){

	out<<"("<<v.first<<", "<<v.second<<")";return out;}

const int MAX_N = 200010;

const int MAX_V = 100010;

const double eps = 1e-6;

const ll mod = 1000000007;

const int inf = 1 << 29;

const ll linf = 1LL << 60;

const double PI = 3.14159265358979323846;

///////////////////////////////////////////////////////////////////////////////////////////////////





typedef pair <int, int> P;

 

const ll MOD=1000000007;

 

ll power(ll x, ll k, ll p){

    if (k==0) return 1;

    else if (k==1) return x%p;

	ll res = power(x, k / 2, p);

	if (k%2==0) return (res * res)%p;

    else return (res * res % p * x)%p;

}

 

ll dp[2][1010];

 

void solve() {

    ll N, M;

    cin >> N >>M;

    

    if (M==1){

        cout << 1<< endl;

    }

    else {

        int a=0;

        ll b=2;

        while (b<=100010){

            if (M%b==0){

                dp[0][a]=b;

                while (M%b==0){

                    M/=b;

                    dp[1][a]++;

                }

                a++;

            }

            b++;

        }

		if(M != 1) {

			dp[0][a] = M;

			dp[1][a] = 1;

			a++;

		}

        ll ans=1;

        if (a==0){

            cout << N << endl;

        }

        else {

            rep(i, 0, a){

                ll p=1;

                ll q=1;

                for (ll j=N ;j< dp[1][i]+N; j++){

                    p*=j;

                    p%=MOD;

                }

                for(ll j=1; j<dp[1][i]+1; j++){

                    q*=j;

                    q%=MOD;

                }

                ans*=(p*power(q, MOD-2, MOD)) % MOD;

                ans%=MOD;

            }

            cout << ans << endl;

        }

    }

}



int main() {

#ifndef LOCAL

	ios::sync_with_stdio(false);

    cin.tie(0);

#endif

    cout << fixed;

	cout.precision(20);

	srand((unsigned int)time(NULL));

#ifdef LOCAL

	//freopen("in.txt", "wt", stdout); //for tester

    freopen("in.txt", "rt", stdin);

#endif	

	solve();

#ifdef LOCAL

    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

#endif

	return 0;

}




// Fastest known solution for this problem: