#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 5;
long long fac[2 * maxn];
int q, x, y;
long long mul(long long n, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * n % mod;
    n = n * n % mod;
    k >>= 1;
  }
  return res;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i < 2 * maxn; i++) fac[i] = (fac[i - 1] * i) % mod;
}
long long C(long long n, long long m) {
  return (fac[n] * mul(fac[m], mod - 2)) % mod * mul(fac[n - m], mod - 2) % mod;
}
int main() {
  init();
  cin >> q;
  while (q--) {
    scanf("%d%d", &x, &y);
    int c = 0;
    long long ans = mul(2, y - 1);
    for (int i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        int t = 0;
        while (x % i == 0) {
          t++;
          x /= i;
        }
        ans = (ans * C(t + y - 1, t)) % mod;
      }
    }
    if (x > 1) ans = (ans * y) % mod;
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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; 
  * <image>. 



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 Ai ≠ Bi. Since the answer can be very large, print it modulo 109 + 7.

Input

The first line contains one integer q (1 ≤ q ≤ 105) — the number of testcases to solve.

Then q lines follow, each containing two integers xi and yi (1 ≤ xi, yi ≤ 106). Each of these lines represents a testcase.

Output

Print q integers. i-th integer has to be equal to the number of yi-factorizations of xi modulo 109 + 7.

Example

Input

2
6 3
4 2


Output

36
6

Note

In the second testcase of the example there are six y-factorizations:

  * { - 4, - 1}; 
  * { - 2, - 2}; 
  * { - 1, - 4}; 
  * {1, 4}; 
  * {2, 2}; 
  * {4, 1}. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 5;
long long fac[2 * maxn];
int q, x, y;
long long mul(long long n, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * n % mod;
    n = n * n % mod;
    k >>= 1;
  }
  return res;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i < 2 * maxn; i++) fac[i] = (fac[i - 1] * i) % mod;
}
long long C(long long n, long long m) {
  return (fac[n] * mul(fac[m], mod - 2)) % mod * mul(fac[n - m], mod - 2) % mod;
}
int main() {
  init();
  cin >> q;
  while (q--) {
    scanf("%d%d", &x, &y);
    int c = 0;
    long long ans = mul(2, y - 1);
    for (int i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        int t = 0;
        while (x % i == 0) {
          t++;
          x /= i;
        }
        ans = (ans * C(t + y - 1, t)) % mod;
      }
    }
    if (x > 1) ans = (ans * y) % mod;
    printf("%d\n", ans);
  }
  return 0;
}

```