#include <bits/stdc++.h>
const int mod = 1e9 + 7;
using namespace std;
long long jie[200005];
long long qpow(long long a, long long b) {
  long long sum = 1, t = a;
  while (b) {
    if (b & 1) sum *= t;
    sum %= mod;
    t = t * t;
    t %= mod;
    b >>= 1;
  }
  return sum;
}
long long C(long long n, long long m) {
  return jie[n] * qpow(jie[m], mod - 2) % mod * qpow(jie[n - m], mod - 2) % mod;
}
int T;
int main() {
  cin >> T;
  jie[0] = 1;
  for (int i = 1; i <= 200000; i++) {
    jie[i] = jie[i - 1] * i;
    jie[i] %= mod;
  }
  while (T--) {
    int n, k;
    cin >> n >> k;
    k;
    long long sum = 0, all = 0;
    long long ans = 0;
    for (int i = 0; i <= n; i += 2) sum += C(n, i), sum %= mod;
    for (int i = 0; i <= n; i++) all += C(n, i), all %= mod;
    if (k == 0) {
      cout << 1 << endl;
      continue;
    }
    if (n % 2) {
      cout << qpow(sum + 1, k) << endl;
    } else {
      for (int i = 1; i <= k + 1; i++) {
        ans += qpow(sum - 1, max(0, i - 1)) * qpow(all, max(k - i, 0));
        ans %= mod;
      }
      cout << ans << endl;
    }
  }
  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>
const int mod = 1e9 + 7;
using namespace std;
long long jie[200005];
long long qpow(long long a, long long b) {
  long long sum = 1, t = a;
  while (b) {
    if (b & 1) sum *= t;
    sum %= mod;
    t = t * t;
    t %= mod;
    b >>= 1;
  }
  return sum;
}
long long C(long long n, long long m) {
  return jie[n] * qpow(jie[m], mod - 2) % mod * qpow(jie[n - m], mod - 2) % mod;
}
int T;
int main() {
  cin >> T;
  jie[0] = 1;
  for (int i = 1; i <= 200000; i++) {
    jie[i] = jie[i - 1] * i;
    jie[i] %= mod;
  }
  while (T--) {
    int n, k;
    cin >> n >> k;
    k;
    long long sum = 0, all = 0;
    long long ans = 0;
    for (int i = 0; i <= n; i += 2) sum += C(n, i), sum %= mod;
    for (int i = 0; i <= n; i++) all += C(n, i), all %= mod;
    if (k == 0) {
      cout << 1 << endl;
      continue;
    }
    if (n % 2) {
      cout << qpow(sum + 1, k) << endl;
    } else {
      for (int i = 1; i <= k + 1; i++) {
        ans += qpow(sum - 1, max(0, i - 1)) * qpow(all, max(k - i, 0));
        ans %= mod;
      }
      cout << ans << endl;
    }
  }
  return 0;
}

```