Problem:
Score : 400 points

Problem Statement

You are given positive integers N and M.

How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.

Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^9

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.

Sample Input 1

2 6

Sample Output 1

4

Four sequences satisfy the condition: \{a_1, a_2\} = \{1, 6\}, \{2, 3\}, \{3, 2\} and \{6, 1\}.

Sample Input 2

3 12

Sample Output 2

18

Sample Input 3

100000 1000000000

Sample Output 3

957870001

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//file_name:ABC110_B.cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts()
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("YES");}else{puts("NO");}
const int MX = 200005;


    // 整数スキャン（複数）
    /*
    int x;
    scanf("%d",&x);
    int y;
    scanf("%d",&y);
    int z;
    scanf("%d",&y);

    // n個の整数のスキャン
    /*
    ll a[n] = {};
    rep(i,n){
        scanf("%lld",&a[i]);
    }
    */

    // 文字列スキャン
    /*
    string s;
    cin >> s;
    */


int main() {
    ll md = 1000000007;
    const ll mod = (ll)1e9 + 7;
    ll n;
    scanf("%lld",&n);
    ll m;
    scanf("%lld",&m);
    if(m==1){
        cout << m << endl;
        return 0;
    }

    ll soinsuu[100][2];
    rep(i,100){
        rep(j,2){
            soinsuu[i][j]=0;
        }
    }
    ll count = 2;
    ll flag = 0;
    ll now = 0;
    while(count<100000&&m>1){
        if(m%count==0){
            soinsuu[now][0] = count;
            soinsuu[now][1]++;
            flag++;
            m /= count;
        }else{
            if(flag!=0){
                now++;
                flag=0;
                count++;
            }else{
                count++;
            }
        }
    }
    if(soinsuu[0][0]==0||n==1){
        cout << n << endl;
        return 0;
    }

    count = 0;
    ll ans = 1;
    ll anss[100];
    while(soinsuu[count][0]!=0){
        rep(i,100){
            anss[i]=1;
        }
        rep(i,soinsuu[count][1]){
            anss[i] = (n-1+soinsuu[count][1]-i);
            //anss[i] %= md;
        }
        rep(i,soinsuu[count][1]){
            rep(j,100){
                if(anss[j]%(i+1)==0){
                    anss[j] /= (i+1);
                    break;
                }
            }
        }
        rep(i,100){
            ans *= anss[i];
            ans = ans % mod;
        }
        count++;
    }
    //ans = ans % mod;
    cout << ans << endl;
    return 0;
}