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.

def primeFactors(n):
    import math
    ans=[]
    while n % 2 == 0: 
        ans.append(2), 
        n = n / 2

    for i in range(3,int(math.sqrt(n))+1,2): 
          
         
        while n % i== 0: 
            ans.append(int(i)) 
            n = n / i 
              
     
    if n > 2: 
        ans.append(int(n))
    #print(ans)
    return list(set(ans))
    
    
def PowerOFPINnfactorial(n, p): 
  
    # initializing answer 
    ans = 0
  
    # initializing 
    temp = p 
  
    # loop until temp<=n 
    while (temp <= n) : 
  
        # add number of numbers  
        # divisible by n 
        ans += n // temp 
  
        # each time multiply  
        # temp by p 
        temp = temp * p 
      
    return int(ans)
    
    
def power(x, y, p) : 
    res = 1     # Initialize result 
  
    # Update x if it is more 
    # than or equal to p 
    x = x % p  
  
    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 
    
    
bc=list(map(int, input().split()))
b=bc[0]
c=bc[1]
primelist=primeFactors(b)
ans=1
for i in primelist:
    temp=PowerOFPINnfactorial(c,i)
    ans=ans*(power(i,temp,1000000007))
    
print((ans)%1000000007)