#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  int flag = 0;
  if (n % 2 == 1) flag = 1;
  long long eq = 1, ans = 0;
  for (int i = 1; i <= k; i++) {
    if (flag == 1) {
      eq *= binpow(2, n - 1, mod) + 1;
      eq %= mod;
    } else {
      long long gr;
      if (i == k)
        gr = 1;
      else
        gr = binpow(binpow(2, n, mod) % mod, k - i, mod);
      gr %= mod;
      gr *= eq;
      gr %= mod;
      ans += gr;
      eq *= binpow(2, n - 1, mod) - 1;
      eq += mod;
      eq %= mod;
    }
  }
  ans += eq;
  ans %= mod;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(10);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please provide a cpp coded solution to the problem described below:

Moamen and Ezzat are playing a game. They create an array a of n non-negative integers where every element is less than 2^k.

Moamen wins if a_1  \&  a_2  \&  a_3  \&  …  \&  a_n ≥ a_1 ⊕ a_2 ⊕ a_3 ⊕ … ⊕ a_n.

Here \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND), and ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Please calculate the number of winning for Moamen arrays a.

As the result may be very large, print the value modulo 1 000 000 007 (10^9 + 7).

Input

The first line contains a single integer t (1 ≤ t ≤ 5)— the number of test cases. 

Each test case consists of one line containing two integers n and k (1 ≤ n≤ 2⋅ 10^5, 0 ≤ k ≤ 2⋅ 10^5).

Output

For each test case, print a single value — the number of different arrays that Moamen wins with.

Print the result modulo 1 000 000 007 (10^9 + 7).

Example

Input


3
3 1
2 1
4 0


Output


5
2
1

Note

In the first example, n = 3, k = 1. As a result, all the possible arrays are [0,0,0], [0,0,1], [0,1,0], [1,0,0], [1,1,0], [0,1,1], [1,0,1], and [1,1,1].

Moamen wins in only 5 of them: [0,0,0], [1,1,0], [0,1,1], [1,0,1], and [1,1,1].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  int flag = 0;
  if (n % 2 == 1) flag = 1;
  long long eq = 1, ans = 0;
  for (int i = 1; i <= k; i++) {
    if (flag == 1) {
      eq *= binpow(2, n - 1, mod) + 1;
      eq %= mod;
    } else {
      long long gr;
      if (i == k)
        gr = 1;
      else
        gr = binpow(binpow(2, n, mod) % mod, k - i, mod);
      gr %= mod;
      gr *= eq;
      gr %= mod;
      ans += gr;
      eq *= binpow(2, n - 1, mod) - 1;
      eq += mod;
      eq %= mod;
    }
  }
  ans += eq;
  ans %= mod;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(10);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

```