#include <bits/stdc++.h>
using namespace std;
long long int pow_cal[200001];
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return res % 1000000007;
}
long long int abc(long long int n, long long int k) {
  if (n == 1) return pow_cal[k];
  if (k == 0) return 1;
  if (n & 1) {
    long long int x = (pow_cal[n - 1] + 1) % 1000000007;
    x = power(x, k);
    return x;
  } else {
    if (k == 1) return pow_cal[n - 1];
    long long int x = power(pow_cal[n], k - 1);
    x = x + ((pow_cal[n - 1] - 1) * (abc(n, k - 1))) % 1000000007;
    x %= 1000000007;
    return x;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pow_cal[0] = 1;
  for (int i = 1; i < 200001; i++) {
    pow_cal[i] = (2 * pow_cal[i - 1]) % 1000000007;
  }
  int T;
  cin >> T;
  while (T--) {
    long long int n, k;
    cin >> n >> k;
    cout << abc(n, k) << '\n';
  }
}

### 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 int pow_cal[200001];
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return res % 1000000007;
}
long long int abc(long long int n, long long int k) {
  if (n == 1) return pow_cal[k];
  if (k == 0) return 1;
  if (n & 1) {
    long long int x = (pow_cal[n - 1] + 1) % 1000000007;
    x = power(x, k);
    return x;
  } else {
    if (k == 1) return pow_cal[n - 1];
    long long int x = power(pow_cal[n], k - 1);
    x = x + ((pow_cal[n - 1] - 1) * (abc(n, k - 1))) % 1000000007;
    x %= 1000000007;
    return x;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pow_cal[0] = 1;
  for (int i = 1; i < 200001; i++) {
    pow_cal[i] = (2 * pow_cal[i - 1]) % 1000000007;
  }
  int T;
  cin >> T;
  while (T--) {
    long long int n, k;
    cin >> n >> k;
    cout << abc(n, k) << '\n';
  }
}

```