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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class Primes_and_Multiplication {
	static long mod = 1000000007l;
	private static BigInteger modd;
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		long[] getprimes = getPrimos(1000000);
		
		StringTokenizer st = new StringTokenizer(in.readLine());
		
		long x = Long.parseLong(st.nextToken());
		long n = Long.parseLong(st.nextToken());
		
		long[][] primes = factoresPrimos(x, getprimes);
		
	
		BigInteger N = new BigInteger(n + "");
//		long total = 1;
		BigInteger total = BigInteger.ONE;
		modd = BigInteger.valueOf(mod);
		for(int i = 0; i < primes.length; i++) {
			long p = primes[i][0];
			if(p == 1) break;
			long cur = p;
			BigInteger d = new BigInteger(cur + "");
			while(N.compareTo(d) >= 0) {
				d = d.multiply(BigInteger.valueOf(p));
			}
			d = d.divide(BigInteger.valueOf(p));
			cur = d.longValue();
			long last = 0;
			while(cur != 1) {
				long ocu = n / cur;
				if(ocu == 0) break;
				long m = (ocu - last);
				BigInteger xd = getpot(BigInteger.valueOf(cur), m);
//				long xd = getpot(cur, m);
				
//				total = (total * xd) % mod;
				total = total.multiply(xd).mod(modd);
//				total = ((total * (((ocu-last) * cur) % mod))) % mod;
//				total = total.mod(modd);
				cur /= p;
				last = ocu;
			}
		}
		
		out.println(total);
	
		
		out.close();
	}
	
	private static BigInteger getpot(BigInteger cur, long m) {
		if(m == 1)
			return cur;
		if(m % 2 == 0) {
			BigInteger sol = getpot(cur, m/2);
			return sol.multiply(sol).mod(modd); 
		}
		else {
			BigInteger sol = getpot(cur, (m-1)/2);
			return sol.multiply(sol).mod(modd).multiply(cur).mod(modd); 
		}
	}

	static long [][] factoresPrimos( long x , long [] primos ) {
		List< long []> res = new ArrayList< long []>(); double R =( long )(Math. sqrt ( x )+1+1e-5); long p , c ;
		for ( int i =0, n = primos . length ; i < n &&( p = primos [ i ])<= R ; i ++) {
		for ( c =0; x % p ==0; x /= p , c ++); if ( c >0) res .add( new long []{ p , c });
		}
		if ( x >1) res .add( new long []{ x ,1});
		return res .toArray( new long [0][]);
		}
	
	static long [] getPrimos( int cota ) {
		boolean [] arr = new boolean [ cota + 1];
		long [] res = new long [ cota ];
		arr [0] = arr [1] = true ;
		int K = 0; for ( int i = 0; i < arr . length ; i ++)
		if (! arr [ i ]) {
		res [ K ++] = i ;
		for ( int j = 2 * i ; j < arr . length ; j += i ) arr [ j ] = true ;
		}
		return Arrays. copyOfRange ( res , 0, K );
		}
}

