#include <bits/stdc++.h>
using namespace std;
int n, k, T, mod = 1e9 + 7;
long long ans = 0, inv[200010], frac[200010], res;
long long qpow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long C(int x, int y) { return frac[x] * inv[y] % mod * inv[x - y] % mod; }
int main() {
  cin >> T;
  frac[0] = inv[0] = 1;
  for (int i = 1; i <= 200000; i++) frac[i] = frac[i - 1] * i % mod;
  for (int i = 1; i <= 200000; i++) inv[i] = qpow(frac[i], mod - 2);
  while (T--) {
    cin >> n >> k;
    ans = res = 0;
    for (int i = 0; i < n; i += 2) ans = (ans + C(n, i)) % mod;
    if (n & 1) {
      ans++;
      printf("%lld\n", qpow(ans, k) % mod);
    } else {
      res = qpow(ans, k);
      for (int i = 1; i <= k; i++)
        res = (res + qpow(qpow(2, n), k - i) * qpow(ans, i - 1)) % mod;
      printf("%lld\n", res);
    }
  }
}

### Prompt

Please formulate 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;
int n, k, T, mod = 1e9 + 7;
long long ans = 0, inv[200010], frac[200010], res;
long long qpow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long C(int x, int y) { return frac[x] * inv[y] % mod * inv[x - y] % mod; }
int main() {
  cin >> T;
  frac[0] = inv[0] = 1;
  for (int i = 1; i <= 200000; i++) frac[i] = frac[i - 1] * i % mod;
  for (int i = 1; i <= 200000; i++) inv[i] = qpow(frac[i], mod - 2);
  while (T--) {
    cin >> n >> k;
    ans = res = 0;
    for (int i = 0; i < n; i += 2) ans = (ans + C(n, i)) % mod;
    if (n & 1) {
      ans++;
      printf("%lld\n", qpow(ans, k) % mod);
    } else {
      res = qpow(ans, k);
      for (int i = 1; i <= k; i++)
        res = (res + qpow(qpow(2, n), k - i) * qpow(ans, i - 1)) % mod;
      printf("%lld\n", res);
    }
  }
}

```