#include <bits/stdc++.h>
using namespace std;
void _io() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
const int MAXN = 200005;
const long long mod = 1e9 + 7;
long long inp1, inp2, inp3;
long long T, x, y;
long long p[MAXN], num[MAXN], top;
namespace Comb {
const int maxc = 2000000 + 5;
int f[maxc], inv[maxc], finv[maxc];
void init() {
  inv[1] = 1;
  for (int i = 2; i < maxc; i++)
    inv[i] = (mod - mod / i) * 1ll * inv[mod % i] % mod;
  f[0] = finv[0] = 1;
  for (int i = 1; i < maxc; i++) {
    f[i] = f[i - 1] * 1ll * i % mod;
    finv[i] = finv[i - 1] * 1ll * inv[i] % mod;
  }
}
int C(int n, int m) {
  if (m < 0 || m > n) return 0;
  return f[n] * 1ll * finv[n - m] % mod * finv[m] % mod;
}
int S(int n, int m) {
  if (n == 0 && m == 0) return 1;
  return C(m + n - 1, n - 1);
}
}  // namespace Comb
using Comb::C;
long long qpow(long long a, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = (ret * a) % mod;
    n >>= 1;
    a = (a * a) % mod;
  }
  return ret;
}
int main() {
  Comb::init();
  long long ans = 0;
  cin >> T;
  while (T--) {
    long long ans = 1;
    top = 0;
    cin >> x >> y;
    for (long long i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        p[++top] = i;
        num[top] = 0;
        while (x % i == 0) {
          num[top]++;
          x /= i;
        }
      }
    }
    if (x > 1) {
      p[++top] = x;
      num[top] = 1;
    }
    for (long long i = 1; i <= top; i++) {
      ans = (ans * C(y + num[i] - 1, y - 1)) % mod;
    }
    ans = (ans * qpow(2, y - 1)) % mod;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
void _io() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
const int MAXN = 200005;
const long long mod = 1e9 + 7;
long long inp1, inp2, inp3;
long long T, x, y;
long long p[MAXN], num[MAXN], top;
namespace Comb {
const int maxc = 2000000 + 5;
int f[maxc], inv[maxc], finv[maxc];
void init() {
  inv[1] = 1;
  for (int i = 2; i < maxc; i++)
    inv[i] = (mod - mod / i) * 1ll * inv[mod % i] % mod;
  f[0] = finv[0] = 1;
  for (int i = 1; i < maxc; i++) {
    f[i] = f[i - 1] * 1ll * i % mod;
    finv[i] = finv[i - 1] * 1ll * inv[i] % mod;
  }
}
int C(int n, int m) {
  if (m < 0 || m > n) return 0;
  return f[n] * 1ll * finv[n - m] % mod * finv[m] % mod;
}
int S(int n, int m) {
  if (n == 0 && m == 0) return 1;
  return C(m + n - 1, n - 1);
}
}  // namespace Comb
using Comb::C;
long long qpow(long long a, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = (ret * a) % mod;
    n >>= 1;
    a = (a * a) % mod;
  }
  return ret;
}
int main() {
  Comb::init();
  long long ans = 0;
  cin >> T;
  while (T--) {
    long long ans = 1;
    top = 0;
    cin >> x >> y;
    for (long long i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        p[++top] = i;
        num[top] = 0;
        while (x % i == 0) {
          num[top]++;
          x /= i;
        }
      }
    }
    if (x > 1) {
      p[++top] = x;
      num[top] = 1;
    }
    for (long long i = 1; i <= top; i++) {
      ans = (ans * C(y + num[i] - 1, y - 1)) % mod;
    }
    ans = (ans * qpow(2, y - 1)) % mod;
    cout << ans << endl;
  }
  return 0;
}

```