#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
const double pi = acos(-1.0);
inline void read(long long &a) {
  a = 0;
  long long d = 1;
  char c;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') d = -1;
  a = a * 10 + c - 48;
  while (c = getchar(), c >= '0' && c <= '9') a = a * 10 + c - 48;
  a *= d;
}
long long fpm(long long b, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * b) % M;
    p >>= 1;
    b = (b * b) % M;
  }
  return ans;
}
long long cc(long long n, long long m) {
  long long ans = 1;
  m = min(m, n - m);
  for (long long i = 0; i < m; i++) ans = (ans * (n - i) / (1 + i)) % M;
  return ans;
}
signed main() {
  long long t;
  read(t);
  while (t--) {
    long long n, k;
    read(n);
    read(k);
    if (k == 0) {
      printf("1\n");
      continue;
    }
    if (n % 2 == 0) {
      long long a = fpm(2, n);
      long long b = fpm(2, n - 1) - 1;
      long long ans = fpm(b, k);
      for (long long i = 1; i <= k; i++)
        ans = (ans + fpm(a, k - i) * fpm(b, i - 1)) % M;
      printf("%lld\n", ans);
    } else {
      long long b = 1 + fpm(2, n - 1);
      long long ans = fpm(b, k);
      printf("%lld\n", ans);
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 M = 1e9 + 7;
const double pi = acos(-1.0);
inline void read(long long &a) {
  a = 0;
  long long d = 1;
  char c;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') d = -1;
  a = a * 10 + c - 48;
  while (c = getchar(), c >= '0' && c <= '9') a = a * 10 + c - 48;
  a *= d;
}
long long fpm(long long b, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * b) % M;
    p >>= 1;
    b = (b * b) % M;
  }
  return ans;
}
long long cc(long long n, long long m) {
  long long ans = 1;
  m = min(m, n - m);
  for (long long i = 0; i < m; i++) ans = (ans * (n - i) / (1 + i)) % M;
  return ans;
}
signed main() {
  long long t;
  read(t);
  while (t--) {
    long long n, k;
    read(n);
    read(k);
    if (k == 0) {
      printf("1\n");
      continue;
    }
    if (n % 2 == 0) {
      long long a = fpm(2, n);
      long long b = fpm(2, n - 1) - 1;
      long long ans = fpm(b, k);
      for (long long i = 1; i <= k; i++)
        ans = (ans + fpm(a, k - i) * fpm(b, i - 1)) % M;
      printf("%lld\n", ans);
    } else {
      long long b = 1 + fpm(2, n - 1);
      long long ans = fpm(b, k);
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```