#include <bits/stdc++.h>
const int N = 2e6;
const int p = 1e9 + 7;
using namespace std;
int T, x, y, fac[N + 5], inv[N + 5], ans;
int mypow(int a, int x) {
  int s = 1;
  for (; x; x& 1 ? s = 1ll * s * a % p : 0, a = 1ll * a * a % p, x >>= 1)
    ;
  return s;
}
int C(int n, int m) { return 1ll * fac[n] * inv[n - m] % p * inv[m] % p; }
int main() {
  scanf("%d", &T);
  fac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % p;
  inv[1] = 1;
  for (int i = 2; i <= N; i++) inv[i] = 1ll * (p - p / i) * inv[p % i] % p;
  inv[0] = 1;
  for (int i = 1; i <= N; i++) inv[i] = 1ll * inv[i - 1] * inv[i] % p;
  while (T--) {
    scanf("%d%d", &x, &y);
    ans = mypow(2, y - 1);
    int xx = x;
    for (int i = 2; i * i <= xx; i++)
      if (xx % i == 0) {
        int cnt = 0;
        while (xx % i == 0) {
          xx /= i;
          cnt++;
        }
        ans = 1ll * ans * C(cnt + y - 1, y - 1) % p;
      }
    if (xx != 1) ans = 1ll * ans * y % p;
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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>
const int N = 2e6;
const int p = 1e9 + 7;
using namespace std;
int T, x, y, fac[N + 5], inv[N + 5], ans;
int mypow(int a, int x) {
  int s = 1;
  for (; x; x& 1 ? s = 1ll * s * a % p : 0, a = 1ll * a * a % p, x >>= 1)
    ;
  return s;
}
int C(int n, int m) { return 1ll * fac[n] * inv[n - m] % p * inv[m] % p; }
int main() {
  scanf("%d", &T);
  fac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % p;
  inv[1] = 1;
  for (int i = 2; i <= N; i++) inv[i] = 1ll * (p - p / i) * inv[p % i] % p;
  inv[0] = 1;
  for (int i = 1; i <= N; i++) inv[i] = 1ll * inv[i - 1] * inv[i] % p;
  while (T--) {
    scanf("%d%d", &x, &y);
    ans = mypow(2, y - 1);
    int xx = x;
    for (int i = 2; i * i <= xx; i++)
      if (xx % i == 0) {
        int cnt = 0;
        while (xx % i == 0) {
          xx /= i;
          cnt++;
        }
        ans = 1ll * ans * C(cnt + y - 1, y - 1) % p;
      }
    if (xx != 1) ans = 1ll * ans * y % p;
    printf("%d\n", ans);
  }
  return 0;
}

```