#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
const long long int maxn = 2e5 + 16, md = 1e9 + 7, inf = 2e16;
long long int gcd(long long int a, long long int b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int tav(long long int n, long long int k) {
  long long int res = 1;
  while (k > 0) {
    if (k & 1) {
      res *= n;
      res %= md;
    }
    n *= n;
    n %= md;
    k >>= 1;
  }
  return res;
}
long long int tv[maxn];
void solve() {
  long long int n, k, cur = 1, ans = 0, w, d;
  cin >> n >> k;
  if (n & 1) {
    w = 0;
  } else {
    w = 1;
  }
  if (n & 1) {
    d = 1 + tv[n - 1];
  } else {
    d = tv[n - 1] - 1;
  }
  for (long long int i = k - 1; i >= 0; i--) {
    long long int h = 1ll * w * cur * tav(2, 1ll * i * n) % md;
    ans += h;
    cur *= d;
    cur %= md;
  }
  ans += cur;
  ans %= md;
  cout << ans << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  tv[0] = 1;
  for (long long int i = 1; i < maxn; i++) {
    tv[i] = (tv[i - 1] << 1);
    if (tv[i] >= md) tv[i] -= md;
  }
  long long int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
#pragma GCC optimize("Ofast")
const long long int maxn = 2e5 + 16, md = 1e9 + 7, inf = 2e16;
long long int gcd(long long int a, long long int b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int tav(long long int n, long long int k) {
  long long int res = 1;
  while (k > 0) {
    if (k & 1) {
      res *= n;
      res %= md;
    }
    n *= n;
    n %= md;
    k >>= 1;
  }
  return res;
}
long long int tv[maxn];
void solve() {
  long long int n, k, cur = 1, ans = 0, w, d;
  cin >> n >> k;
  if (n & 1) {
    w = 0;
  } else {
    w = 1;
  }
  if (n & 1) {
    d = 1 + tv[n - 1];
  } else {
    d = tv[n - 1] - 1;
  }
  for (long long int i = k - 1; i >= 0; i--) {
    long long int h = 1ll * w * cur * tav(2, 1ll * i * n) % md;
    ans += h;
    cur *= d;
    cur %= md;
  }
  ans += cur;
  ans %= md;
  cout << ans << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  tv[0] = 1;
  for (long long int i = 1; i < maxn; i++) {
    tv[i] = (tv[i - 1] << 1);
    if (tv[i] >= md) tv[i] -= md;
  }
  long long int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```