#include <bits/stdc++.h>
using namespace std;
int const MAX = 2e5 + 5;
int const MOD = 1e9 + 7;
int powers[MAX];
void precalc() {
  powers[0] = 1;
  for (int i = 1; i < MAX; ++i) {
    powers[i] = powers[i - 1] * 2;
    powers[i] %= MOD;
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  int sol = 1;
  if (k == 0) {
    cout << 1 << "\n";
    return;
  }
  if (n % 2 == 1) {
    for (int i = 0; i < k; ++i) {
      sol = ((long long)sol * (powers[n - 1] + 1)) % MOD;
    }
    cout << sol << "\n";
    return;
  }
  sol = 0;
  int pre[k];
  pre[0] = powers[n];
  for (int i = 1; i < k; ++i) {
    pre[i] = (((long long)powers[n] * pre[i - 1]) % MOD);
  }
  int aft[k];
  aft[0] = powers[n - 1] - 1;
  for (int i = 1; i < k; ++i) {
    aft[i] = (((long long)(powers[n - 1] - 1) * aft[i - 1]) % MOD);
  }
  sol += aft[k - 1];
  for (int i = 0; i < k; ++i) {
    long long total = 1;
    if (i > 0) {
      total *= pre[i - 1];
    }
    if (i < k - 1) {
      total *= aft[k - 2 - i];
    }
    total %= MOD;
    sol += total;
    sol %= MOD;
  }
  cout << sol << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  precalc();
  for (int i = 0; i < t; ++i) {
    solve();
  }
}

### Prompt

Generate a CPP solution to the following problem:

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;
int const MAX = 2e5 + 5;
int const MOD = 1e9 + 7;
int powers[MAX];
void precalc() {
  powers[0] = 1;
  for (int i = 1; i < MAX; ++i) {
    powers[i] = powers[i - 1] * 2;
    powers[i] %= MOD;
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  int sol = 1;
  if (k == 0) {
    cout << 1 << "\n";
    return;
  }
  if (n % 2 == 1) {
    for (int i = 0; i < k; ++i) {
      sol = ((long long)sol * (powers[n - 1] + 1)) % MOD;
    }
    cout << sol << "\n";
    return;
  }
  sol = 0;
  int pre[k];
  pre[0] = powers[n];
  for (int i = 1; i < k; ++i) {
    pre[i] = (((long long)powers[n] * pre[i - 1]) % MOD);
  }
  int aft[k];
  aft[0] = powers[n - 1] - 1;
  for (int i = 1; i < k; ++i) {
    aft[i] = (((long long)(powers[n - 1] - 1) * aft[i - 1]) % MOD);
  }
  sol += aft[k - 1];
  for (int i = 0; i < k; ++i) {
    long long total = 1;
    if (i > 0) {
      total *= pre[i - 1];
    }
    if (i < k - 1) {
      total *= aft[k - 2 - i];
    }
    total %= MOD;
    sol += total;
    sol %= MOD;
  }
  cout << sol << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  precalc();
  for (int i = 0; i < t; ++i) {
    solve();
  }
}

```