#include <bits/stdc++.h>
using namespace std;
long long n, k;
signed main() {
  long long T;
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld%lld", &n, &k);
    long long w = 0, l = 0, b = 1;
    long long ta = 1, tb = 1, tc = 1;
    for (long long i = 0; i < n; i++) {
      tc <<= 1;
      tc %= 1000000007;
    }
    for (long long i = 1; i < n; i++) {
      ta <<= 1;
      ta %= 1000000007;
      tb <<= 1;
      tb %= 1000000007;
    }
    if (n & 1) {
      ta--;
      if (ta < 0) ta += 1000000007;
      tb--;
      if (tb < 0) tb += 1000000007;
    } else {
      tb -= 2;
      if (tb < 0) tb += 1000000007;
    }
    while (k--) {
      if (n & 1) {
        w *= tc;
        w %= 1000000007;
        b = (2 + tb) * b;
        b %= 1000000007;
      } else {
        w *= tc;
        w %= 1000000007;
        w += b;
        w %= 1000000007;
        b = (1 + tb) * b;
        b %= 1000000007;
      }
    }
    printf("%lld\n", (w + b) % 1000000007);
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
long long n, k;
signed main() {
  long long T;
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld%lld", &n, &k);
    long long w = 0, l = 0, b = 1;
    long long ta = 1, tb = 1, tc = 1;
    for (long long i = 0; i < n; i++) {
      tc <<= 1;
      tc %= 1000000007;
    }
    for (long long i = 1; i < n; i++) {
      ta <<= 1;
      ta %= 1000000007;
      tb <<= 1;
      tb %= 1000000007;
    }
    if (n & 1) {
      ta--;
      if (ta < 0) ta += 1000000007;
      tb--;
      if (tb < 0) tb += 1000000007;
    } else {
      tb -= 2;
      if (tb < 0) tb += 1000000007;
    }
    while (k--) {
      if (n & 1) {
        w *= tc;
        w %= 1000000007;
        b = (2 + tb) * b;
        b %= 1000000007;
      } else {
        w *= tc;
        w %= 1000000007;
        w += b;
        w %= 1000000007;
        b = (1 + tb) * b;
        b %= 1000000007;
      }
    }
    printf("%lld\n", (w + b) % 1000000007);
  }
  return 0;
}

```