#include <bits/stdc++.h>
using namespace std;
const int _MAX = 1e+9 + 7;
int _pow(int n, int __pow, vector<int>* _mem = nullptr) {
  int res = 1;
  if (_mem != nullptr) _mem->emplace_back(res);
  while (__pow--) {
    res = ((long long)res * n) % _MAX;
    if (_mem != nullptr) _mem->emplace_back(res);
  }
  return res;
}
int solve(int n, int k) {
  if (!k) return 1;
  int _f = _pow(2, n - 1), _g = _pow(2, n) - 1;
  if (n % 2) return _pow(((long long)_f + 1) % _MAX, k);
  _f--;
  vector<int> _f_mem, _g_mem;
  int res = _pow(_f, k, &_f_mem);
  _pow((1 + (long long)_g) % _MAX, k - 1, &_g_mem);
  for (int i = 0; i <= k - 1; i++)
    res = ((((long long)_f_mem[i] * _g_mem[k - 1 - i]) % _MAX) + res) % _MAX;
  return res;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    cout << solve(n, k) << endl;
  }
}

### Prompt

Your task is to create 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;
const int _MAX = 1e+9 + 7;
int _pow(int n, int __pow, vector<int>* _mem = nullptr) {
  int res = 1;
  if (_mem != nullptr) _mem->emplace_back(res);
  while (__pow--) {
    res = ((long long)res * n) % _MAX;
    if (_mem != nullptr) _mem->emplace_back(res);
  }
  return res;
}
int solve(int n, int k) {
  if (!k) return 1;
  int _f = _pow(2, n - 1), _g = _pow(2, n) - 1;
  if (n % 2) return _pow(((long long)_f + 1) % _MAX, k);
  _f--;
  vector<int> _f_mem, _g_mem;
  int res = _pow(_f, k, &_f_mem);
  _pow((1 + (long long)_g) % _MAX, k - 1, &_g_mem);
  for (int i = 0; i <= k - 1; i++)
    res = ((((long long)_f_mem[i] * _g_mem[k - 1 - i]) % _MAX) + res) % _MAX;
  return res;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    cout << solve(n, k) << endl;
  }
}

```