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.util.*;
import java.io.*;
public class PrimesandMultiplication
{
	public static void main(String[] f)
	{
		long x,n;
		long h=1;
		long h1=1000000007l;
		Scanner sc=new Scanner(System.in);
		x=sc.nextLong();
		n=sc.nextLong();
		HashMap<Long,Long> hm=primefactors(x);
		for(Map.Entry<Long,Long> g:hm.entrySet())
		{
			long key=g.getKey();
			long val=g.getValue();
			long val1=0;
			long m=n;
			long a=key;
			while(m>=a)
			{
			long o=(m/a);
			h*=(modPow(key,o,h1)%h1);
			h%=h1;
			if((m/a)<key)
				break;
			a*=key;
			}
		}
		System.out.println(h);
	}
	public static HashMap<Long,Long> primefactors(long x)
	{
		long j=2,count1=0;
		HashMap<Long,Long> hm=new HashMap<>();
		while(x%j==0)
		{
			count1++;
			x/=j;
		}
		if(count1>0)
		hm.put(j,count1);
		for(j=3;j<=Math.sqrt(x);j+=2)
		{
			count1=0;
			while(x%j==0)
			{
				count1++;
				x/=j;
			}
			if(count1>0)
			{
				hm.put(j,count1);
			}
		}
		if(x>2)
			hm.put(x,1l);
		return hm;
	}
	public static long modPow(long base,long expo,long mod)
	{
		if(expo==0)
			return 1;
		if(expo%2==0)
		{
			long b=modPow(base,(expo/2),mod);
			b%=mod;
			return ((b*b)%mod);
		}
		else
		{
			long b=modPow(base,(expo-1),mod);
			b%=mod;
			return (((base%mod)*b)%mod);
		}
	}
}