#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 10, p = 1e9 + 7;
inline long long ksm(long long a, long long b, const long long p) {
  long long c = 1;
  for (a %= p; b; a = a * a % p, b >>= 1)
    if (b & 1) c = c * a % p;
  return c;
}
void solve() {
  long long n, k, res = 0;
  cin >> n >> k;
  long long bs = ksm(2, n - 1, p), bs2 = bs * 2 % p;
  if (n & 1)
    cout << ksm(bs + 1, k, p) << '\n';
  else {
    for (long long x = k; x; --x) {
      long long a = ksm(bs + p - 1, k - x, p), b = ksm(bs2, x - 1, p);
      res = (res + a * b % p) % p;
    }
    res = (res + ksm(bs + p - 1, k, p)) % p;
    cout << res << '\n';
  }
}
signed main() {
  cout.tie(0)->sync_with_stdio(0);
  long long _;
  for (cin >> _; _--;) solve();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long N = 2e5 + 10, p = 1e9 + 7;
inline long long ksm(long long a, long long b, const long long p) {
  long long c = 1;
  for (a %= p; b; a = a * a % p, b >>= 1)
    if (b & 1) c = c * a % p;
  return c;
}
void solve() {
  long long n, k, res = 0;
  cin >> n >> k;
  long long bs = ksm(2, n - 1, p), bs2 = bs * 2 % p;
  if (n & 1)
    cout << ksm(bs + 1, k, p) << '\n';
  else {
    for (long long x = k; x; --x) {
      long long a = ksm(bs + p - 1, k - x, p), b = ksm(bs2, x - 1, p);
      res = (res + a * b % p) % p;
    }
    res = (res + ksm(bs + p - 1, k, p)) % p;
    cout << res << '\n';
  }
}
signed main() {
  cout.tie(0)->sync_with_stdio(0);
  long long _;
  for (cin >> _; _--;) solve();
  return 0;
}

```