#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, y = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') y = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * y;
}
long long b[100005];
long long c[100005];
const int mod = 1e9 + 7;
long long qpow(long long x, long long y) {
  long long res = 1, m = x;
  while (y) {
    if (y & 1) res = (res * m) % mod;
    m = (m * m) % mod;
    y >>= 1;
  }
  return res;
}
int main() {
  int T = read();
  while (T--) {
    long long n = read(), k = read();
    if (n & 1) {
      long long ans = qpow(2, n - 1);
      ans = (ans + 1) % mod;
      ans = qpow(ans, k);
      printf("%lld\n", ans);
    } else {
      long long ans = 1;
      long long m = 1;
      for (int i = 1; i <= k; i++) {
        ans = ((ans * (qpow(2, n - 1) - 1)) % mod + m) % mod;
        m = (m * qpow(2, n)) % mod;
      }
      ans = ((ans % mod) + mod) % mod;
      printf("%lld\n", ans);
    }
  }
  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;
long long read() {
  long long x = 0, y = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') y = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * y;
}
long long b[100005];
long long c[100005];
const int mod = 1e9 + 7;
long long qpow(long long x, long long y) {
  long long res = 1, m = x;
  while (y) {
    if (y & 1) res = (res * m) % mod;
    m = (m * m) % mod;
    y >>= 1;
  }
  return res;
}
int main() {
  int T = read();
  while (T--) {
    long long n = read(), k = read();
    if (n & 1) {
      long long ans = qpow(2, n - 1);
      ans = (ans + 1) % mod;
      ans = qpow(ans, k);
      printf("%lld\n", ans);
    } else {
      long long ans = 1;
      long long m = 1;
      for (int i = 1; i <= k; i++) {
        ans = ((ans * (qpow(2, n - 1) - 1)) % mod + m) % mod;
        m = (m * qpow(2, n)) % mod;
      }
      ans = ((ans % mod) + mod) % mod;
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```