#include <bits/stdc++.h>
using lint = long long;
constexpr lint maxn = 2e5 + 5;
constexpr lint inf = 1e9;
constexpr lint mod = 1e9 + 7;
using namespace std;
using pii = pair<lint, lint>;
lint t, n, k;
lint dp[maxn][2];
lint exp(lint a, lint b) {
  a %= mod;
  lint ret = 1;
  while (b) {
    if (b & 1) ret *= a;
    a *= a;
    ret %= mod;
    a %= mod;
    b >>= 1;
  }
  return ret;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    if (k == 0) {
      cout << 1 << '\n';
      continue;
    }
    if (n % 2 == 0) {
      lint comb = exp(2, n - 1) - 1;
      dp[0][1] = 1;
      dp[0][0] = comb;
      for (lint i = 1; i < k; i++) {
        lint prev = (dp[i - 1][0] + dp[i - 1][1]) % mod;
        dp[i][0] = (prev * comb) % mod;
        dp[i][1] = exp(exp(2, mod), n * i / mod) * exp(2, n * i % mod) % mod;
      }
    } else {
      lint comb = exp(2, n - 1);
      dp[0][1] = 1;
      dp[0][0] = comb;
      for (lint i = 1; i < k; i++) {
        lint prev = (dp[i - 1][0] + dp[i - 1][1]) % mod;
        dp[i][0] = (prev * comb) % mod;
        dp[i][1] = prev % mod;
      }
    }
    cout << (dp[k - 1][0] + dp[k - 1][1]) % mod << '\n';
  }
}

### 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 lint = long long;
constexpr lint maxn = 2e5 + 5;
constexpr lint inf = 1e9;
constexpr lint mod = 1e9 + 7;
using namespace std;
using pii = pair<lint, lint>;
lint t, n, k;
lint dp[maxn][2];
lint exp(lint a, lint b) {
  a %= mod;
  lint ret = 1;
  while (b) {
    if (b & 1) ret *= a;
    a *= a;
    ret %= mod;
    a %= mod;
    b >>= 1;
  }
  return ret;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    if (k == 0) {
      cout << 1 << '\n';
      continue;
    }
    if (n % 2 == 0) {
      lint comb = exp(2, n - 1) - 1;
      dp[0][1] = 1;
      dp[0][0] = comb;
      for (lint i = 1; i < k; i++) {
        lint prev = (dp[i - 1][0] + dp[i - 1][1]) % mod;
        dp[i][0] = (prev * comb) % mod;
        dp[i][1] = exp(exp(2, mod), n * i / mod) * exp(2, n * i % mod) % mod;
      }
    } else {
      lint comb = exp(2, n - 1);
      dp[0][1] = 1;
      dp[0][0] = comb;
      for (lint i = 1; i < k; i++) {
        lint prev = (dp[i - 1][0] + dp[i - 1][1]) % mod;
        dp[i][0] = (prev * comb) % mod;
        dp[i][1] = prev % mod;
      }
    }
    cout << (dp[k - 1][0] + dp[k - 1][1]) % mod << '\n';
  }
}

```