#include <bits/stdc++.h>
using namespace std;
long long dp[1001];
bool ready[1001];
bool used[1001];
long double dist(pair<long long, long long> p1, pair<long long, long long> p2) {
  return sqrt((p2.first - p1.first) * (p2.first - p1.first) +
              (p2.second - p1.second) * (p2.second - p1.second));
}
string repeat(long long t, string s) {
  string ans;
  for (long long i = 0; i < t; i++) {
    ans += s;
  }
  return ans;
}
long long bin_mod_exp(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  if (b == 1) {
    return a % 1000000007;
  }
  a %= 1000000007;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = ((res % 1000000007) * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (n == 1) {
    cout << bin_mod_exp(2LL, k) << endl;
    return;
  }
  if (!k) {
    cout << 1 << endl;
    return;
  }
  if (n % 2 == 0) {
    long long a = bin_mod_exp(bin_mod_exp(2LL, n - 1) + 1, 1000000007 - 2);
    long long ans = ((bin_mod_exp(2LL, n * k) -
                      bin_mod_exp(bin_mod_exp(2LL, n - 1) - 1, k)) *
                         a +
                     bin_mod_exp(bin_mod_exp(2LL, n - 1) - 1, k)) %
                    1000000007;
    while (ans < 0) {
      ans += 1000000007;
    }
    cout << ans << endl;
  } else {
    cout << bin_mod_exp(bin_mod_exp(2LL, n - 1) + 1, k) << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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>
using namespace std;
long long dp[1001];
bool ready[1001];
bool used[1001];
long double dist(pair<long long, long long> p1, pair<long long, long long> p2) {
  return sqrt((p2.first - p1.first) * (p2.first - p1.first) +
              (p2.second - p1.second) * (p2.second - p1.second));
}
string repeat(long long t, string s) {
  string ans;
  for (long long i = 0; i < t; i++) {
    ans += s;
  }
  return ans;
}
long long bin_mod_exp(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  if (b == 1) {
    return a % 1000000007;
  }
  a %= 1000000007;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = ((res % 1000000007) * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (n == 1) {
    cout << bin_mod_exp(2LL, k) << endl;
    return;
  }
  if (!k) {
    cout << 1 << endl;
    return;
  }
  if (n % 2 == 0) {
    long long a = bin_mod_exp(bin_mod_exp(2LL, n - 1) + 1, 1000000007 - 2);
    long long ans = ((bin_mod_exp(2LL, n * k) -
                      bin_mod_exp(bin_mod_exp(2LL, n - 1) - 1, k)) *
                         a +
                     bin_mod_exp(bin_mod_exp(2LL, n - 1) - 1, k)) %
                    1000000007;
    while (ans < 0) {
      ans += 1000000007;
    }
    cout << ans << endl;
  } else {
    cout << bin_mod_exp(bin_mod_exp(2LL, n - 1) + 1, k) << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```