#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 3e5 + 5;
int f[N], inv[N], XOR, dp[N], n, k;
int pow(int a, int b) {
  int res = 1;
  while (b) {
    if (b % 2) {
      res = 1ll * res * a % MOD;
    }
    a = 1ll * a * a % MOD;
    b /= 2;
  }
  return res;
}
int pow(int a, long long b) {
  int res = 1;
  while (b) {
    if (b % 2) {
      res = 1ll * res * a % MOD;
    }
    a = 1ll * a * a % MOD;
    b /= 2;
  }
  return res;
}
void precompute() {
  f[1] = 1;
  for (int i = 2; i < N; ++i) {
    f[i] = 1ll * f[i - 1] * i % MOD;
  }
  inv[N - 1] = pow(f[N - 1], MOD - 2);
  for (int i = N - 2; i >= 0; --i) {
    inv[i] = 1ll * inv[i + 1] * (i + 1) % MOD;
  }
}
int nCr(int x, int y) {
  return x - y >= 0 ? 1ll * f[x] * inv[y] % MOD * inv[x - y] % MOD : 0;
}
int go(int bit) {
  if (bit == k) {
    return 1;
  }
  if (dp[bit] != -1) {
    return dp[bit];
  }
  int cnt = 0;
  if (n % 2 == 0) {
    cnt = (cnt + pow(2, 1ll * n * (k - bit - 1))) % MOD;
  }
  if (n % 2) {
    cnt = (cnt + go(bit + 1)) % MOD;
  }
  cnt = (cnt + 1ll * XOR * go(bit + 1)) % MOD;
  return dp[bit] = cnt;
}
void solve() {
  cin >> n >> k;
  memset(dp, -1, sizeof(dp));
  XOR = 0;
  for (int i = 0; i < n; i += 2) {
    XOR = (XOR + nCr(n, i)) % MOD;
  }
  cout << go(0) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  precompute();
  int t = 1;
  cin >> t;
  int i = 1;
  while (t--) {
    solve();
  }
  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 MOD = 1e9 + 7;
const int N = 3e5 + 5;
int f[N], inv[N], XOR, dp[N], n, k;
int pow(int a, int b) {
  int res = 1;
  while (b) {
    if (b % 2) {
      res = 1ll * res * a % MOD;
    }
    a = 1ll * a * a % MOD;
    b /= 2;
  }
  return res;
}
int pow(int a, long long b) {
  int res = 1;
  while (b) {
    if (b % 2) {
      res = 1ll * res * a % MOD;
    }
    a = 1ll * a * a % MOD;
    b /= 2;
  }
  return res;
}
void precompute() {
  f[1] = 1;
  for (int i = 2; i < N; ++i) {
    f[i] = 1ll * f[i - 1] * i % MOD;
  }
  inv[N - 1] = pow(f[N - 1], MOD - 2);
  for (int i = N - 2; i >= 0; --i) {
    inv[i] = 1ll * inv[i + 1] * (i + 1) % MOD;
  }
}
int nCr(int x, int y) {
  return x - y >= 0 ? 1ll * f[x] * inv[y] % MOD * inv[x - y] % MOD : 0;
}
int go(int bit) {
  if (bit == k) {
    return 1;
  }
  if (dp[bit] != -1) {
    return dp[bit];
  }
  int cnt = 0;
  if (n % 2 == 0) {
    cnt = (cnt + pow(2, 1ll * n * (k - bit - 1))) % MOD;
  }
  if (n % 2) {
    cnt = (cnt + go(bit + 1)) % MOD;
  }
  cnt = (cnt + 1ll * XOR * go(bit + 1)) % MOD;
  return dp[bit] = cnt;
}
void solve() {
  cin >> n >> k;
  memset(dp, -1, sizeof(dp));
  XOR = 0;
  for (int i = 0; i < n; i += 2) {
    XOR = (XOR + nCr(n, i)) % MOD;
  }
  cout << go(0) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  precompute();
  int t = 1;
  cin >> t;
  int i = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```