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.
# Python3 program to print prime factors 
# and their powers. 
import math 

# Function to calculate all the prime 
# factors and count of every prime factor 
def factorize(n): 
    count = 0
    # count the number of 
    li=[]
    while ((n % 2 > 0) == False): 
        
        # equivalent to n = n / 2; 
        n >>= 1; 
        count += 1; 

    # if 2 divides it 
    if (count > 0): 
        li.append(2)

    # check for all the possible 
    # numbers that can divide it 
    for i in range(3, int(math.sqrt(n)) + 1): 
        count = 0; 
        while (n % i == 0): 
            count += 1; 
            n = int(n / i); 
        if (count > 0): 
            li.append(i)
        i += 2; 

    # if n at the end is a prime number. 
    if (n > 2): 
        li.append(n)
    return li

# Driver Code 
x,n=map(int,input().split())
s=factorize(x)
ans=1
m=10**9+7
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 
ns=len(s)
for j in range(ns):
    p=0
    c=s[j]
    i=1
    while(c<=n):
        p+=n//c
        i+=1
        c=s[j]**i
    ans=((ans)*(power(s[j],p,m)))%m
print(ans)