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 math


def power(x, y, p) : 
    res = 1     # Initialize result 
  
    # Update x if it is more 
    # than or equal to p 
    x = x % p  
      
    if (x == 0) : 
        return 0
  
    while (y > 0) : 
          
        # If y is odd, multiply 
        # x with result 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
  
        # y must be even now 
        y = y >> 1      # y = y/2 
        x = (x * x) % p 
          
    return res


def isPrime(n) : 
    if (n <= 1) : 
        return False
    if (n <= 3) : 
        return True
    if (n % 2 == 0 or n % 3 == 0) : 
        return False

    i = 5
    while(i * i <= n) : 
        if (n % i == 0 or n % (i + 2) == 0) : 
            return False
        i = i + 6
    return True

def prime(n): 
    res = set()
    i = 1
    while i <= math.sqrt(n): 
        if (n % i == 0) : 
            if (n / i == i) and isPrime(i): 
                res.add(i) 
            else : 
                if isPrime(i):
                    res.add(i) 
                if isPrime(n//i):
                    res.add(n//i) 
        i = i + 1
    return res


def solve(x,n):
    res = 1
    P = prime(x)
    for y in P:
        if y>n:
            continue
        temp =[]
        cur = y
        while True:
            nos = n//cur
            if(nos==0):
                break
            temp.append(nos)
            cur*=y
        for i in range(len(temp)-1,-1,-1):
            for j in range(i-1,-1,-1):
                temp[j]-=temp[i]
        MOD = int(1000000007)
        for i in range(len(temp)):
            num = int(power(y,i+1,92233720368547758071))
            cur = power(num,temp[i],MOD)
            res*=cur
            res%=MOD
    return res
x,n = map(int,input().split())
print(solve(x,n))
