#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10, Mod = 1e9 + 7;
int t, n, k;
long long Pow(long long a, long long b) {
  a %= Mod;
  long long ans = 1;
  while (b) {
    if (b & 1) ans *= a, ans %= Mod;
    a *= a, a %= Mod;
    b >>= 1;
  }
  return ans;
}
long long jie[MAXN], ni[MAXN];
void getall() {
  jie[0] = 1, ni[0] = 1;
  for (int i = 1; i < MAXN; ++i) jie[i] = (jie[i - 1] * i) % Mod;
  ni[MAXN - 1] = Pow(jie[MAXN - 1], Mod - 2);
  for (int i = MAXN - 2; i >= 1; --i) ni[i] = (ni[i + 1] * (i + 1)) % Mod;
}
long long calc(int a) {
  long long ans = jie[n];
  ans *= ni[a], ans %= Mod;
  ans *= ni[n - a], ans %= Mod;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> t;
  getall();
  while (t--) {
    cin >> n >> k;
    if (k == 0) {
      cout << 1 << endl;
      continue;
    }
    long long ans = 0, tmp1, tmp2, tt = Pow(2, k);
    if (n & 1) {
      tmp1 = 1;
      for (int i = 0; i <= n; i += 2) tmp1 += calc(i), tmp1 %= Mod;
      cout << Pow(tmp1, k) << endl;
    } else {
      tmp1 = 0, tmp2 = 1, ans = 0;
      for (int i = 0; i < n; i += 2) tmp1 += calc(i), tmp1 %= Mod;
      ans = Pow(tmp1, k);
      for (int i = 1; i <= k; ++i) {
        ans += tmp2 * Pow(2, 1ll * (k - i) * n);
        ans %= Mod;
        tmp2 *= tmp1;
        tmp2 %= Mod;
      }
      cout << ans << endl;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 MAXN = 2e5 + 10, Mod = 1e9 + 7;
int t, n, k;
long long Pow(long long a, long long b) {
  a %= Mod;
  long long ans = 1;
  while (b) {
    if (b & 1) ans *= a, ans %= Mod;
    a *= a, a %= Mod;
    b >>= 1;
  }
  return ans;
}
long long jie[MAXN], ni[MAXN];
void getall() {
  jie[0] = 1, ni[0] = 1;
  for (int i = 1; i < MAXN; ++i) jie[i] = (jie[i - 1] * i) % Mod;
  ni[MAXN - 1] = Pow(jie[MAXN - 1], Mod - 2);
  for (int i = MAXN - 2; i >= 1; --i) ni[i] = (ni[i + 1] * (i + 1)) % Mod;
}
long long calc(int a) {
  long long ans = jie[n];
  ans *= ni[a], ans %= Mod;
  ans *= ni[n - a], ans %= Mod;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> t;
  getall();
  while (t--) {
    cin >> n >> k;
    if (k == 0) {
      cout << 1 << endl;
      continue;
    }
    long long ans = 0, tmp1, tmp2, tt = Pow(2, k);
    if (n & 1) {
      tmp1 = 1;
      for (int i = 0; i <= n; i += 2) tmp1 += calc(i), tmp1 %= Mod;
      cout << Pow(tmp1, k) << endl;
    } else {
      tmp1 = 0, tmp2 = 1, ans = 0;
      for (int i = 0; i < n; i += 2) tmp1 += calc(i), tmp1 %= Mod;
      ans = Pow(tmp1, k);
      for (int i = 1; i <= k; ++i) {
        ans += tmp2 * Pow(2, 1ll * (k - i) * n);
        ans %= Mod;
        tmp2 *= tmp1;
        tmp2 %= Mod;
      }
      cout << ans << endl;
    }
  }
  return 0;
}

```