#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int mod_pow(int b, int p) {
  int ans = 1;
  while (p) {
    if (p & 1) {
      ans = (1ll * ans * b) % mod;
      p--;
    }
    b = (1ll * b * b) % mod;
    p /= 2;
  }
  return ans;
}
const int N = (2e5 + 5);
int fac[N], inv_fac[N];
void calc_fac() {
  fac[0] = 1;
  inv_fac[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = (1ll * fac[i - 1] * i) % mod;
    inv_fac[i] = mod_pow(fac[i], mod - 2);
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << 1;
    cout << '\n';
    return;
  }
  int e = 0;
  for (int i = 0; i < n; i += 2) {
    e = (e + 1ll * fac[n] * inv_fac[n - i] % mod * inv_fac[i] % mod) % mod;
  }
  if (n & 1) e = (e + 1) % mod;
  if (n & 1) {
    int ans = mod_pow(e, k);
    cout << ans;
    cout << '\n';
    return;
  }
  int p = mod_pow(2, n);
  int pp = mod_pow(p, k);
  int ans = 0;
  int left = 1;
  int right = pp;
  for (int i = 0; i < k; i++) {
    right = 1ll * right * mod_pow(p, mod - 2) % mod;
    ans = (ans + 1ll * left * right) % mod;
    left = (1ll * left * e) % mod;
  }
  ans = (ans + 1ll * left) % mod;
  cout << ans;
  cout << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  calc_fac();
  int tc;
  cin >> tc;
  while (tc--) solve();
  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;
const int mod = 1e9 + 7;
int mod_pow(int b, int p) {
  int ans = 1;
  while (p) {
    if (p & 1) {
      ans = (1ll * ans * b) % mod;
      p--;
    }
    b = (1ll * b * b) % mod;
    p /= 2;
  }
  return ans;
}
const int N = (2e5 + 5);
int fac[N], inv_fac[N];
void calc_fac() {
  fac[0] = 1;
  inv_fac[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = (1ll * fac[i - 1] * i) % mod;
    inv_fac[i] = mod_pow(fac[i], mod - 2);
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << 1;
    cout << '\n';
    return;
  }
  int e = 0;
  for (int i = 0; i < n; i += 2) {
    e = (e + 1ll * fac[n] * inv_fac[n - i] % mod * inv_fac[i] % mod) % mod;
  }
  if (n & 1) e = (e + 1) % mod;
  if (n & 1) {
    int ans = mod_pow(e, k);
    cout << ans;
    cout << '\n';
    return;
  }
  int p = mod_pow(2, n);
  int pp = mod_pow(p, k);
  int ans = 0;
  int left = 1;
  int right = pp;
  for (int i = 0; i < k; i++) {
    right = 1ll * right * mod_pow(p, mod - 2) % mod;
    ans = (ans + 1ll * left * right) % mod;
    left = (1ll * left * e) % mod;
  }
  ans = (ans + 1ll * left) % mod;
  cout << ans;
  cout << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  calc_fac();
  int tc;
  cin >> tc;
  while (tc--) solve();
  return 0;
}

```