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
from collections import defaultdict, deque, Counter
from decimal import *
from heapq import heapify, heappop, heappush
import math
import random
import string
from copy import deepcopy
from itertools import combinations, permutations, product
from operator import mul, itemgetter
from functools import reduce
from bisect import bisect_left, bisect_right

def input():
    return sys.stdin.readline().rstrip()
def getN():
    return int(input())
def getNM():
    return map(int, input().split())
def getList():
    return list(map(int, input().split()))
def getListGraph():
    return list(map(lambda x:int(x) - 1, input().split()))
def getArray(intn):
    return [int(input()) for i in range(intn)]

mod = 10 ** 9 + 7
MOD = 998244353
# sys.setrecursionlimit(10000000)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')
inf = float('inf')
eps = 10 ** (-12)
dy = [0, 1, 0, -1]
dx = [1, 0, -1, 0]

#############
# Main Code #
#############

# 素因数分解
def prime_factorize(n):
    divisors = []
    # 27(2 * 2 * 7)の7を出すためにtemp使う
    temp = n
    for i in range(2, int(math.sqrt(n)) + 1):
        if temp % i == 0:
            cnt = 0
            while temp % i == 0:
                cnt += 1
                # 素因数を見つけるたびにtempを割っていく
                temp //= i
            divisors.append([i, cnt])
    if temp != 1:
        divisors.append([temp, 1])
    if divisors == []:
        divisors.append([n, 1])

    return divisors

lim = 10 ** 6 + 1
fact = [1, 1]
factinv = [1, 1]
inv = [0, 1]

for i in range(2, lim + 1):
    fact.append((fact[-1] * i) % mod)
    inv.append((-inv[mod % i] * (mod // i)) % mod)
    # 累計
    factinv.append((factinv[-1] * inv[-1]) % mod)

def cmb(n, r):
    if (r < 0) or (n < r):
        return 0
    r = min(r, n - r)
    return fact[n] * factinv[r] * factinv[n - r] % mod

"""
まずxを素因数分解
36 = 2^2 * 3^2
どこに因数を配分するか　またマイナスはどこか？　写像１２相
1つ目に2を置き、3つ目にまた2を置く　その逆と同じでダブル

玉を区別しない、箱は区別するなので重複組み合わせ
k個の玉、Y個の箱（Y-1個の仕切り)

マイナスを2n個ずつ置く
前もってdpしておく
dp[i][0] i番目終了時点でプラスのもの dp[i][1] i番目でマイナスのもの
"""

dp = [[0 for i in range(2)] for i in range(10 ** 6 + 7)]
dp[0][0] = 1
for i in range(1, 10 ** 6 + 7):
    dp[i][0] = dp[i][1] = dp[i - 1][0] + dp[i - 1][1]
    dp[i][0] %= mod
    dp[i][1] %= mod

T = getN()
for _ in range(T):
    X, Y = getNM()
    ans = dp[Y][0]
    for _, k in prime_factorize(X):
        ans *= cmb(k + Y - 1, k)
        ans %= mod
    print(ans)