#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
long long sz(long long T, long long pot) {
  if (pot == 1) {
    return T;
  }
  if (pot % 2 == 1)
    return ((sz((T * T) % M, pot / 2) * T) + M) % M;
  else
    return (sz((T * T) % M, pot / 2) + M) % M;
}
long long OP[200005];
long long IV[200005];
int main() {
  for (int i = 1; i < 200005; i++) {
    IV[i] = sz(i, M - 2);
  }
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    long long S = 0;
    long long R = 0;
    long long Q = 1;
    for (int j = 1; j <= (n - 1) / 2; j++) {
      Q *= (n - 2 * j + 2);
      Q = (Q + M) % M;
      Q *= IV[2 * j - 1];
      Q = (Q + M) % M;
      Q *= (n - 2 * j + 1);
      Q = (Q + M) % M;
      Q *= IV[2 * j];
      Q = (Q + M) % M;
      R += Q;
      R = (R + M) % M;
    }
    if (n % 2 == 0) {
      R++;
      R = (R + M) % M;
    }
    OP[0] = 1;
    for (int i = 1; i <= k; i++) {
      OP[i] = OP[i - 1] * R;
      OP[i] = (OP[i] + M) % M;
    }
    if (n % 2 == 1) {
      long long W = 1;
      long long dwa = 1;
      for (int i = 0; i <= k; i++) {
        if (i == 0) {
          if (k != 0)
            S = OP[k];
          else
            S = 1;
        } else {
          if (i != k) {
            W *= (k - i + 1);
            W = (W + M) % M;
            W *= IV[i];
            W = (W + M) % M;
            dwa *= 2;
            dwa = (dwa + M) % M;
            long long vi = dwa;
            vi *= W;
            vi = (vi + M) % M;
            vi *= OP[k - i];
            vi = (vi + M) % M;
            S += vi;
            S = (S + M) % M;
          } else {
            dwa *= 2;
            dwa = (dwa + M) % M;
            S += dwa;
          }
          S = (S + M) % M;
        }
      }
    } else {
      long long dwa = sz(2, n);
      long long Dwa = 1;
      for (int i = 0; i <= k; i++) {
        if (i == 0) {
          if (k != 0)
            S = OP[k];
          else
            S = 1;
        } else if (i == 1) {
          S += OP[k - 1];
        } else {
          Dwa *= dwa;
          Dwa = (Dwa + M) % M;
          long long vi = Dwa;
          vi *= OP[k - i];
          vi = (vi + M) % M;
          S += vi;
          S = (S + M) % M;
        }
      }
    }
    cout << S << "\n";
  }
  return 0;
}

### Prompt

Please formulate a CPP solution 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;
const int M = 1e9 + 7;
long long sz(long long T, long long pot) {
  if (pot == 1) {
    return T;
  }
  if (pot % 2 == 1)
    return ((sz((T * T) % M, pot / 2) * T) + M) % M;
  else
    return (sz((T * T) % M, pot / 2) + M) % M;
}
long long OP[200005];
long long IV[200005];
int main() {
  for (int i = 1; i < 200005; i++) {
    IV[i] = sz(i, M - 2);
  }
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    long long S = 0;
    long long R = 0;
    long long Q = 1;
    for (int j = 1; j <= (n - 1) / 2; j++) {
      Q *= (n - 2 * j + 2);
      Q = (Q + M) % M;
      Q *= IV[2 * j - 1];
      Q = (Q + M) % M;
      Q *= (n - 2 * j + 1);
      Q = (Q + M) % M;
      Q *= IV[2 * j];
      Q = (Q + M) % M;
      R += Q;
      R = (R + M) % M;
    }
    if (n % 2 == 0) {
      R++;
      R = (R + M) % M;
    }
    OP[0] = 1;
    for (int i = 1; i <= k; i++) {
      OP[i] = OP[i - 1] * R;
      OP[i] = (OP[i] + M) % M;
    }
    if (n % 2 == 1) {
      long long W = 1;
      long long dwa = 1;
      for (int i = 0; i <= k; i++) {
        if (i == 0) {
          if (k != 0)
            S = OP[k];
          else
            S = 1;
        } else {
          if (i != k) {
            W *= (k - i + 1);
            W = (W + M) % M;
            W *= IV[i];
            W = (W + M) % M;
            dwa *= 2;
            dwa = (dwa + M) % M;
            long long vi = dwa;
            vi *= W;
            vi = (vi + M) % M;
            vi *= OP[k - i];
            vi = (vi + M) % M;
            S += vi;
            S = (S + M) % M;
          } else {
            dwa *= 2;
            dwa = (dwa + M) % M;
            S += dwa;
          }
          S = (S + M) % M;
        }
      }
    } else {
      long long dwa = sz(2, n);
      long long Dwa = 1;
      for (int i = 0; i <= k; i++) {
        if (i == 0) {
          if (k != 0)
            S = OP[k];
          else
            S = 1;
        } else if (i == 1) {
          S += OP[k - 1];
        } else {
          Dwa *= dwa;
          Dwa = (Dwa + M) % M;
          long long vi = Dwa;
          vi *= OP[k - i];
          vi = (vi + M) % M;
          S += vi;
          S = (S + M) % M;
        }
      }
    }
    cout << S << "\n";
  }
  return 0;
}

```