#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
using namespace std;
const long long N = 1e6 + 5;
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
long long t, x, k, fact[10 * N];
bitset<1000005> pv;
long long spf[N];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < N; i++)
    if (!pv[i]) {
      spf[i] = i;
      for (long long j = i + i; j < N; j += i) {
        pv[j] = true;
        if (!spf[j]) spf[j] = i;
      }
    }
}
void fac() {
  fact[0] = 1;
  for (long long i = 1; i < 10 * N; i++)
    fact[i] = i * fact[i - 1], fact[i] %= mod;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  sieve(), fac();
  cin >> t;
  while (t--) {
    cin >> x >> k;
    long long s = 1;
    long long a = x;
    map<long long, long long> M;
    while (a > 1) M[spf[a]]++, a /= spf[a];
    for (auto it : M) {
      long long ki = it.second;
      s *= fact[ki + k - 1], s %= mod;
      s *= modexpo(fact[k - 1], mod - 2), s %= mod;
      s *= modexpo(fact[ki], mod - 2), s %= mod;
    }
    s *= modexpo(2, k - 1), s %= mod;
    cout << s << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
#pragma GCC optimize("-O3")
using namespace std;
const long long N = 1e6 + 5;
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
long long t, x, k, fact[10 * N];
bitset<1000005> pv;
long long spf[N];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < N; i++)
    if (!pv[i]) {
      spf[i] = i;
      for (long long j = i + i; j < N; j += i) {
        pv[j] = true;
        if (!spf[j]) spf[j] = i;
      }
    }
}
void fac() {
  fact[0] = 1;
  for (long long i = 1; i < 10 * N; i++)
    fact[i] = i * fact[i - 1], fact[i] %= mod;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  sieve(), fac();
  cin >> t;
  while (t--) {
    cin >> x >> k;
    long long s = 1;
    long long a = x;
    map<long long, long long> M;
    while (a > 1) M[spf[a]]++, a /= spf[a];
    for (auto it : M) {
      long long ki = it.second;
      s *= fact[ki + k - 1], s %= mod;
      s *= modexpo(fact[k - 1], mod - 2), s %= mod;
      s *= modexpo(fact[ki], mod - 2), s %= mod;
    }
    s *= modexpo(2, k - 1), s %= mod;
    cout << s << "\n";
  }
  return 0;
}

```