#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n, k;
const int N = 200020;
long long dp[N];
int qpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return res;
}
void solve() {
  cin >> n >> k;
  long long ans = qpow(2, n - 1);
  if (n & 1) {
    int t1 = (qpow(2, n - 1) + 1) % mod;
    cout << qpow(t1, k) << endl;
  } else {
    long long res = 0;
    for (int i = 0; i < k; ++i) {
      res = (res +
             1ll * qpow(ans * 2 % mod, i) * qpow(ans - 1, k - i - 1) % mod) %
            mod;
    }
    res = (res + qpow(ans - 1, k)) % mod;
    cout << res << endl;
  }
}
int T = 1;
signed main() {
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Please create a solution in cpp 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 mod = 1e9 + 7;
int n, k;
const int N = 200020;
long long dp[N];
int qpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return res;
}
void solve() {
  cin >> n >> k;
  long long ans = qpow(2, n - 1);
  if (n & 1) {
    int t1 = (qpow(2, n - 1) + 1) % mod;
    cout << qpow(t1, k) << endl;
  } else {
    long long res = 0;
    for (int i = 0; i < k; ++i) {
      res = (res +
             1ll * qpow(ans * 2 % mod, i) * qpow(ans - 1, k - i - 1) % mod) %
            mod;
    }
    res = (res + qpow(ans - 1, k)) % mod;
    cout << res << endl;
  }
}
int T = 1;
signed main() {
  cin >> T;
  while (T--) solve();
  return 0;
}

```