You are given two positive integer numbers *x* and *y*. An array *F* is called an *y*-factorization of *x* iff the following conditions are met:
 -  There are *y* elements in *F*, and all of them are integer numbers; -  <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3d08f4583882898cfb5a2c074c0b1c98e3c52a95.png" style="max-width: 100.0%;max-height: 100.0%;"/>. 
You have to count the number of pairwise distinct arrays that are *y*-factorizations of *x*. Two arrays *A* and *B* are considered different iff there exists at least one index *i* (1 ≤ *i* ≤ *y*) such that *A*<sub class="lower-index">*i*</sub> ≠ *B*<sub class="lower-index">*i*</sub>. Since the answer can be very large, print it modulo 10<sup class="upper-index">9</sup> + 7.
The first line contains one integer *q* (1 ≤ *q* ≤ 10<sup class="upper-index">5</sup>) — the number of testcases to solve.

Then *q* lines follow, each containing two integers *x*<sub class="lower-index">*i*</sub> and *y*<sub class="lower-index">*i*</sub> (1 ≤ *x*<sub class="lower-index">*i*</sub>, *y*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">6</sup>). Each of these lines represents a testcase.
In the second testcase of the example there are six *y*-factorizations:
 -  { - 4,  - 1}; -  { - 2,  - 2}; -  { - 1,  - 4}; -  {1, 4}; -  {2, 2}; -  {4, 1}.
2
6 3
4 2
36
6
1
524288 1000000
645043186
1
65536 1000000
928522471
1
5612 11399
215664246
import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

import math
mod = 10**9+7

def main():
    N = 2*10**6
    spf = [-1]*(N+1)
    for i in range(N+1):
        spf[i] = i
    for i in range(2, int(math.sqrt(N))+1):
        if spf[i] == i:
            for j in range(i*2, N+1, i):
                if spf[j] == j:
                    spf[j] = i

    def factorize(n):
        d = {}
        while n != 1:
            p = spf[n]
            if p in d:
                d[p] += 1
            else:
                d[p] = 1
            n //= p
        return d

    fac = [1]*(N+1)
    finv = [1]*(N+1)
    for i in range(N):
        fac[i+1] = fac[i] * (i+1) % mod
    finv[-1] = pow(fac[-1], mod-2, mod)

    for i in reversed(range(N)):
        finv[i] = finv[i+1] * (i+1) % mod

    def cmb1(n, r, mod):
        if r <0 or r > n:
            return 0
        r = min(r, n-r)
        return fac[n] * finv[r] * finv[n-r] % mod

    table = [0]*N
    table[0] = 1
    for i in range(1, N):
        table[i] = table[i-1]*2
        table[i] %= mod

    q = int(input())
    ANS = [0]*q
    for i in range(q):
        x, y = map(int, input().split())
        d = factorize(x)
        #print(d)
        ans = 1
        for k, v in d.items():
            ans *= cmb1(v+y-1, y-1, mod)
            ans %= mod
        ans *= table[y-1]
        ans %= mod
        ANS[i] = ans
    print(*ANS, sep='\n')

if __name__ == '__main__':
    main()
