#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long dp1[200005][2];
long long dp3[200005];
long long f1(long long i, long long j) {
  if (dp1[i][j] != -1) return dp1[i][j];
  if (i == 0) {
    if (j == 0)
      return dp1[i][j] = 1;
    else
      return dp1[i][j] = 0;
  }
  return dp1[i][j] = (f1(i - 1, j) + f1(i - 1, (j + 1) % 2)) % mod;
}
long long f2(vector<long long> &dp2, long long i, long long n) {
  if (i == 0) {
    return dp3[i] = 1;
  }
  if (dp3[i] != -1) return dp3[i];
  long long ans = 0;
  if (n % 2) {
    ans += f2(dp2, i - 1, n);
    ans %= mod;
  } else {
    ans += dp2[i - 1];
    ans %= mod;
  }
  long long d = dp1[n][0];
  if (n % 2 == 0) {
    d = (d - 1 + mod) % mod;
  }
  ans += d * f2(dp2, i - 1, n) % mod;
  ans %= mod;
  return dp3[i] = ans;
}
void solvef() {
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < 200005; i++) dp1[i][0] = dp1[i][1] = -1;
  f1(n, 0);
  if (k == 0) {
    cout << 1 << endl;
    return;
  }
  long long pn = 1;
  for (int i = 0; i < n; i++) pn = pn * 2 % mod;
  vector<long long> dp2(k + 1);
  dp2[0] = 1;
  for (int i = 1; i <= k; i++) {
    dp2[i] = dp2[i - 1] * pn % mod;
  }
  for (int i = 0; i < 200005; i++) dp3[i] = -1;
  long long ans = f2(dp2, k, n);
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solvef();
  return 0;
}

### Prompt

Generate 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;
long long mod = 1e9 + 7;
long long dp1[200005][2];
long long dp3[200005];
long long f1(long long i, long long j) {
  if (dp1[i][j] != -1) return dp1[i][j];
  if (i == 0) {
    if (j == 0)
      return dp1[i][j] = 1;
    else
      return dp1[i][j] = 0;
  }
  return dp1[i][j] = (f1(i - 1, j) + f1(i - 1, (j + 1) % 2)) % mod;
}
long long f2(vector<long long> &dp2, long long i, long long n) {
  if (i == 0) {
    return dp3[i] = 1;
  }
  if (dp3[i] != -1) return dp3[i];
  long long ans = 0;
  if (n % 2) {
    ans += f2(dp2, i - 1, n);
    ans %= mod;
  } else {
    ans += dp2[i - 1];
    ans %= mod;
  }
  long long d = dp1[n][0];
  if (n % 2 == 0) {
    d = (d - 1 + mod) % mod;
  }
  ans += d * f2(dp2, i - 1, n) % mod;
  ans %= mod;
  return dp3[i] = ans;
}
void solvef() {
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < 200005; i++) dp1[i][0] = dp1[i][1] = -1;
  f1(n, 0);
  if (k == 0) {
    cout << 1 << endl;
    return;
  }
  long long pn = 1;
  for (int i = 0; i < n; i++) pn = pn * 2 % mod;
  vector<long long> dp2(k + 1);
  dp2[0] = 1;
  for (int i = 1; i <= k; i++) {
    dp2[i] = dp2[i - 1] * pn % mod;
  }
  for (int i = 0; i < 200005; i++) dp3[i] = -1;
  long long ans = f2(dp2, k, n);
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solvef();
  return 0;
}

```