Let's introduce some definitions that will be needed later.

Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.

Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:

  * g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27), 
  * g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49). 



Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:

  * f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10, 
  * f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63. 



You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.

Input

The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.

Output

Print the answer.

Examples

Input


10 2


Output


2


Input


20190929 1605


Output


363165664


Input


947 987654321987654321


Output


593574252

Note

In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.

In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).

In the third example, be careful about overflow issue.
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main{
    public static final int maxn=(int)1e5+100;
    static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));
    static BufferedReader buff=new BufferedReader(new InputStreamReader(System.in));
    static Scanner sc=new Scanner(System.in);
    static long mod=(long)1e9+7;
    static boolean isP[];
    static int p[],pl;
    static void getP(){
        isP=new boolean[maxn];
        p=new int[maxn];
        for(int i=2;i<maxn;i++){
            if(!isP[i]){
                p[pl++]=i;
                for(int j=2;i*j<maxn;j++){
                    isP[i*j]=true;
                }
            }
        }
    }
    static Vector<Integer> v;
    static void getN(int x){
        v=new Vector<Integer>();
        for(int i=0;i<pl;i++){
            if(x%p[i]==0)v.add(p[i]);
            while(x%p[i]==0)x/=p[i];
        }
        if(x!=1)v.add(x);
    }
    static long pow(long x,long y){
        long ans=1;
        long t=x;
        while(y!=0){
            if(y%2==1)ans*=t;
            t*=t;
            ans%=mod;
            t%=mod;
            y>>=1;
        }
        return ans;
    }
    public static void main(String args[])throws Exception{
        BigInteger a=f(2);
        int x=sc.nextInt();
        long y=sc.nextLong();
        //System.out.println(y);
        long ans=1;
        BigInteger an=BigInteger.ONE;
        getP();
        getN(x);
        for(int i=0;i<v.size();i++){
            int cur=v.get(i);
            BigInteger t=f(cur);
            while(t.compareTo(f(y))<=0){
                BigInteger p=f(y).divide(t);
               // System.out.println(y+" "+t+" "+p);
               // p%=(mod-1);
                an=an.multiply(f(cur).modPow(p,f(mod)));
                //System.out.println(ans+" "+an);
                ans%=mod;
                an=an.mod(f(mod));
                t=t.multiply(f(cur));
            }
        }
        System.out.println(an);
    }
    static BigInteger f(long x){
        return new BigInteger(x+"");
    }
    static String getS()throws Exception{
        in.nextToken();
        return in.sval;
    }
    static int getInt()throws Exception{
        in.nextToken();
        return (int)in.nval;
    }
}