#include <bits/stdc++.h>
using namespace std;
const int N = 2000006;
long long fat[N], lp[N];
long long pot(long long b, long long e) {
  long long ans = 1;
  b %= 1000000007;
  while (e) {
    if (e & 1) ans = (ans * b) % 1000000007;
    b = (b * b) % 1000000007;
    e >>= 1;
  }
  return ans;
}
long long C(int n, int k) {
  return (fat[n] * pot(fat[k] * fat[n - k], 1000000007 - 2)) % 1000000007;
}
long long f(int x, int y) {
  long long ans = 1;
  while (x > 1) {
    int cnt = 0, l = lp[x];
    while (lp[x] == l) cnt++, x /= lp[x];
    ans = (ans * C(y + cnt - 1, cnt)) % 1000000007;
  }
  ans = (ans * pot(2, y - 1)) % 1000000007;
  return ans;
}
int main() {
  fat[0] = 1;
  for (int i = 1; i < N; i++) fat[i] = (i * fat[i - 1]) % 1000000007;
  lp[1] = 1;
  for (int i = 2; i < N; i++)
    if (!lp[i]) {
      for (int j = i; j < N; j += i)
        if (!lp[j]) lp[j] = i;
    }
  int t, x, y;
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &x, &y);
    printf("%lld\n", f(x, y));
  }
  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 N = 2000006;
long long fat[N], lp[N];
long long pot(long long b, long long e) {
  long long ans = 1;
  b %= 1000000007;
  while (e) {
    if (e & 1) ans = (ans * b) % 1000000007;
    b = (b * b) % 1000000007;
    e >>= 1;
  }
  return ans;
}
long long C(int n, int k) {
  return (fat[n] * pot(fat[k] * fat[n - k], 1000000007 - 2)) % 1000000007;
}
long long f(int x, int y) {
  long long ans = 1;
  while (x > 1) {
    int cnt = 0, l = lp[x];
    while (lp[x] == l) cnt++, x /= lp[x];
    ans = (ans * C(y + cnt - 1, cnt)) % 1000000007;
  }
  ans = (ans * pot(2, y - 1)) % 1000000007;
  return ans;
}
int main() {
  fat[0] = 1;
  for (int i = 1; i < N; i++) fat[i] = (i * fat[i - 1]) % 1000000007;
  lp[1] = 1;
  for (int i = 2; i < N; i++)
    if (!lp[i]) {
      for (int j = i; j < N; j += i)
        if (!lp[j]) lp[j] = i;
    }
  int t, x, y;
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &x, &y);
    printf("%lld\n", f(x, y));
  }
  return 0;
}

```