#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int mod = 1e9 + 7;
int n, k, t;
inline long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  t = read();
  while (t--) {
    n = read();
    k = read();
    if (n & 1) {
      long long ans = qpow(2, n - 1) + 1;
      ans = qpow(ans, k);
      printf("%lld\n", ans);
    } else {
      long long temp = (qpow(2, n - 1) - 1 + mod) % mod;
      long long tmp = qpow(2, n);
      long long ans = qpow(temp, k);
      for (int i = 1; i <= k; ++i)
        ans = (ans + (long long)qpow(temp, k - i) * qpow(tmp, i - 1)) % mod;
      printf("%lld\n", ans);
    }
  }
}

### Prompt

In cpp, your task is to solve 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>
#pragma GCC optimize(3)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int mod = 1e9 + 7;
int n, k, t;
inline long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  t = read();
  while (t--) {
    n = read();
    k = read();
    if (n & 1) {
      long long ans = qpow(2, n - 1) + 1;
      ans = qpow(ans, k);
      printf("%lld\n", ans);
    } else {
      long long temp = (qpow(2, n - 1) - 1 + mod) % mod;
      long long tmp = qpow(2, n);
      long long ans = qpow(temp, k);
      for (int i = 1; i <= k; ++i)
        ans = (ans + (long long)qpow(temp, k - i) * qpow(tmp, i - 1)) % mod;
      printf("%lld\n", ans);
    }
  }
}

```