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.util.*; 
//import javafx.util.*; 
import java.math.*;
//import java.lang.*;
public class Main 
{ 
    
    // static int n;static ArrayList<Integer> arr[];
    // static int m;
    // static boolean v[];
   // int ans[];
   // static int arr[];
    static final long oo=(long)1e18;
    static int mod=1000000007;
    public static void main(String[] args) throws IOException { 
    //    Scanner sc=new Scanner(System.in);
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        br = new BufferedReader(new InputStreamReader(System.in));
        long x=nextLong();
        long n=nextLong();
       long copy=x;
        ArrayList<Long> div=new ArrayList<Long>();
        int count=0;
        if(x%2==0){
                div.add(2l);
                count++;
            }
        while (x%2==0) 
        { 

            //System.out.print(2 + " "); 
            x /= 2; 
        } 
  
        // n must be odd at this point.  So we can 
        // skip one element (Note i = i +2) 
        for (long i = 3; i <= Math.sqrt(x); i+= 2) 
        { 
            // While i divides n, print i and divide n 
            if(x%i==0){
                div.add(i);
                count++;
            }
            while (x%i == 0) 
            { 
              //  System.out.print(i + " "); 
                x /= i; 
            } 
        } 
        if(x!=1){
            div.add(x);
        }
       //acc pw.println(count);
        long ans=1;
        for(long val:div){
            long a=1;
            for(long i=1;i<100;i++){
               // long a=fast_pow(val,i,oo);
             //   System.out.println(ans);
                if(a > Long.MAX_VALUE/val){
                    break;
                }
                a=a*val;
               // pw.println(a);
                if(a<0){
                    break;
                }
               // long b=modInverse(a,mod);
                if(a>n){
                    break;
                }

                long num=(n/a);
                ans=(ans*fast_pow(val,num,mod))%mod;

               
            }


        }
        pw.println(ans);
        pw.close();
    }
    
    public static BufferedReader br;
    public static StringTokenizer st;
    public static String next() {
        while (st == null || !st.hasMoreTokens()) {
       try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }
 
    public static Integer nextInt() {
        return Integer.parseInt(next());
    }
    public static Long nextLong() {
        return Long.parseLong(next());
    }
 
    public static Double nextDouble() {
        return Double.parseDouble(next());
    }
    static class Pair implements Comparable<Pair>{
        long x;long y;
        Pair(long x,long y){
            this.x=x;
            this.y=y;
        }
        public int compareTo(Pair a){
            if((x+y)-(a.x+a.y)>0)
                return 1;
            else if((x+y)-(a.x+a.y)==0)
                 return 0;
             else
                return -1;
        }
    }
    // static class Pair{
    //     long x;long y;
    //     Pair(long x,long y){
    //         this.x=x;
    //         this.y=y;
    //     }
    // }
    // static class sorting implements Comparator<Pair>{
    //     public long compare(Pair a,Pair b){
    //         //return (a.y)-(b.y);
    //         return (a.x+a.y)-(b.x+b.y);
    //     }
    // }
    public static int[] na(int n)throws IOException{
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = nextInt();
        return a;
    }
    static class query implements Comparable<query>{
        int l,r,idx,block;
        static int len;
        query(int l,int r,int i){
            this.l=l;
            this.r=r;
            this.idx=i;
            this.block=l/len;
        }   
        public int compareTo(query a){
            return block==a.block?r-a.r:block-a.block;
        }
    }
    static boolean isPrime(int n) { 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
        if (n % 2 == 0 ||  
            n % 3 == 0) 
            return false; 
      
        for (int i = 5; 
                 i * i <= n; i = i + 6) 
            if (n % i == 0 || 
                n % (i + 2) == 0) 
                return false; 
      
        return true; 
    } 
    static int gcd(int a, int b) { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
    // static long modInverse(long a, long m) { 
    //     long g = gcd(a, m); 
    //     if (g != 1) 
    //         return -1; 
    //     else{ 
    //         // If a and m are relatively prime, then modulo inverse 
    //         // is a^(m-2) mode m 
    //         return (power(a, m - 2, m)); 
    //     } 
    // } 
      
    // To compute x^y under modulo m 
    static long power(long x, long y, long m){ 
        if (y == 0) 
            return 1;      
        long p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    }
    static long fast_pow(long base,long n,long M){
        if(n==0)
           return 1;
        if(n==1)
        return base;
        long halfn=fast_pow(base,n/2,M);
        if(n%2==0)
            return ( halfn * halfn ) % M;
        else
            return ( ( ( halfn * halfn ) % M ) * base ) % M;
    }
       static long modInverse(long n,int M){
        return fast_pow(n,M-2,M);
    }
    
} 

