#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mod = 1e9 + 7;
const int nax = 2e5;
int T, n, k;
ll qexp(int a, int b) {
  ll res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = (ll)a * a % mod;
    b >>= 1;
  }
  return res;
}
void solve() {
  cin >> n >> k;
  ll ans = 1, pre = 1;
  if (n & 1) {
    for (int i = k - 1; ~i; i -= 1) ans = ans * (1 + qexp(2, n - 1)) % mod;
  } else {
    ans = 0;
    for (int i = k - 1; ~i; i -= 1) {
      ans = (ans + pre * qexp(qexp(2, n), i) % mod) % mod;
      pre = pre * (qexp(2, n - 1) - 1 + mod) % mod;
    }
    ans = (ans + pre) % mod;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Create a solution in CPP for 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;
using ll = long long;
const int mod = 1e9 + 7;
const int nax = 2e5;
int T, n, k;
ll qexp(int a, int b) {
  ll res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = (ll)a * a % mod;
    b >>= 1;
  }
  return res;
}
void solve() {
  cin >> n >> k;
  ll ans = 1, pre = 1;
  if (n & 1) {
    for (int i = k - 1; ~i; i -= 1) ans = ans * (1 + qexp(2, n - 1)) % mod;
  } else {
    ans = 0;
    for (int i = k - 1; ~i; i -= 1) {
      ans = (ans + pre * qexp(qexp(2, n), i) % mod) % mod;
      pre = pre * (qexp(2, n - 1) - 1 + mod) % mod;
    }
    ans = (ans + pre) % mod;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T--) solve();
  return 0;
}

```