#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const long long MOD = 1e9 + 7;
const int MAX = 2e5 + 10;
long long aux[MAX][2];
long long dp[MAX][2];
int N, K;
long long fexp(long long a, long long e) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * a % MOD;
    a = a * a % MOD;
    e >>= 1;
  }
  return ans;
}
long long solve(int pos, bool won) {
  if (pos == K) return 1;
  long long &ans = dp[pos][won];
  if (ans != -1) return ans;
  ans = 0;
  if (won) {
    ans = (ans + fexp(2, N) * solve(pos + 1, 1) % MOD) % MOD;
  } else {
    ans = (ans + solve(pos + 1, !(N & 1))) % MOD;
    ans = (ans + (aux[N][0] + MOD - !(N & 1)) * solve(pos + 1, 0) % MOD) % MOD;
  }
  return ans;
}
inline void test_case() {
  cin >> N >> K;
  memset(dp, -1, sizeof dp);
  cout << solve(0, 0) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  aux[0][0] = 1;
  for (int i = 1; i < MAX; i++) {
    for (int k : {0, 1}) {
      aux[i][k] = (aux[i - 1][k] + aux[i - 1][not k]) % MOD;
    }
  }
  int number_of_tests;
  cin >> number_of_tests;
  while (number_of_tests--) {
    test_case();
  }
  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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const long long MOD = 1e9 + 7;
const int MAX = 2e5 + 10;
long long aux[MAX][2];
long long dp[MAX][2];
int N, K;
long long fexp(long long a, long long e) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * a % MOD;
    a = a * a % MOD;
    e >>= 1;
  }
  return ans;
}
long long solve(int pos, bool won) {
  if (pos == K) return 1;
  long long &ans = dp[pos][won];
  if (ans != -1) return ans;
  ans = 0;
  if (won) {
    ans = (ans + fexp(2, N) * solve(pos + 1, 1) % MOD) % MOD;
  } else {
    ans = (ans + solve(pos + 1, !(N & 1))) % MOD;
    ans = (ans + (aux[N][0] + MOD - !(N & 1)) * solve(pos + 1, 0) % MOD) % MOD;
  }
  return ans;
}
inline void test_case() {
  cin >> N >> K;
  memset(dp, -1, sizeof dp);
  cout << solve(0, 0) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  aux[0][0] = 1;
  for (int i = 1; i < MAX; i++) {
    for (int k : {0, 1}) {
      aux[i][k] = (aux[i - 1][k] + aux[i - 1][not k]) % MOD;
    }
  }
  int number_of_tests;
  cin >> number_of_tests;
  while (number_of_tests--) {
    test_case();
  }
  return 0;
}

```