#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int MAXN = 2e5 + 100;
int fact[MAXN];
int add(int a, int b) { return (a + b) % mod; }
int mult(int a, int b) { return (a % mod) * (b % mod) % mod; }
int gcd(int a, int b, int &x, int &y) {
  if (!b) {
    x = 1;
    y = 0;
    return 1;
  }
  int x1, y1;
  int ans = gcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return ans;
}
int modInverse(int a) {
  int x, y;
  int g = gcd(a, mod, x, y);
  if (g != 1) {
    return -1;
  } else {
    x = (x % mod + mod) % mod;
    return x;
  }
}
int nCr(int n, int r) {
  if (r > n) return 0;
  if (!r || r == n) return 1;
  return mult(fact[n], modInverse(mult(fact[n - r], fact[r])));
}
int binpow(long long n, long long k) {
  if (k == 0) return 1;
  if (k % 2 == 0)
    return (binpow(n, k / 2) % mod) * (binpow(n, k / 2) % mod) % mod;
  long long tmp = binpow(n, k - 1) % mod;
  return (tmp % mod) * (n % mod) % mod;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  int odd, even = 0;
  for (int i = 0; i < n + 1; i++) {
    if (i % 2 == 1) {
      odd = add(odd, nCr(n, i));
    } else {
      even = add(even, nCr(n, i));
    }
  }
  if (n % 2 == 1) {
    cout << binpow(even + 1, k) << '\n';
    return;
  }
  if (k == 0) {
    cout << 1 << '\n';
    return;
  }
  vector<vector<int>> dp(k + 1, vector<int>(2));
  int all = binpow(2, n);
  dp[1][1] = 1;
  dp[1][0] = even - 1;
  for (int i = 2; i < k + 1; i++) {
    dp[i][0] = mult(dp[i - 1][0], even - 1);
    dp[i][1] = add(dp[i - 1][0], mult(dp[i - 1][1], all));
  }
  cout << add(dp[k][1], dp[k][0]) << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  fact[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    fact[i] = mult(fact[i - 1], i);
  }
  int t = 1;
  cin >> t;
  while (t--) solve();
}

### 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 long long mod = 1e9 + 7;
const int MAXN = 2e5 + 100;
int fact[MAXN];
int add(int a, int b) { return (a + b) % mod; }
int mult(int a, int b) { return (a % mod) * (b % mod) % mod; }
int gcd(int a, int b, int &x, int &y) {
  if (!b) {
    x = 1;
    y = 0;
    return 1;
  }
  int x1, y1;
  int ans = gcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return ans;
}
int modInverse(int a) {
  int x, y;
  int g = gcd(a, mod, x, y);
  if (g != 1) {
    return -1;
  } else {
    x = (x % mod + mod) % mod;
    return x;
  }
}
int nCr(int n, int r) {
  if (r > n) return 0;
  if (!r || r == n) return 1;
  return mult(fact[n], modInverse(mult(fact[n - r], fact[r])));
}
int binpow(long long n, long long k) {
  if (k == 0) return 1;
  if (k % 2 == 0)
    return (binpow(n, k / 2) % mod) * (binpow(n, k / 2) % mod) % mod;
  long long tmp = binpow(n, k - 1) % mod;
  return (tmp % mod) * (n % mod) % mod;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  int odd, even = 0;
  for (int i = 0; i < n + 1; i++) {
    if (i % 2 == 1) {
      odd = add(odd, nCr(n, i));
    } else {
      even = add(even, nCr(n, i));
    }
  }
  if (n % 2 == 1) {
    cout << binpow(even + 1, k) << '\n';
    return;
  }
  if (k == 0) {
    cout << 1 << '\n';
    return;
  }
  vector<vector<int>> dp(k + 1, vector<int>(2));
  int all = binpow(2, n);
  dp[1][1] = 1;
  dp[1][0] = even - 1;
  for (int i = 2; i < k + 1; i++) {
    dp[i][0] = mult(dp[i - 1][0], even - 1);
    dp[i][1] = add(dp[i - 1][0], mult(dp[i - 1][1], all));
  }
  cout << add(dp[k][1], dp[k][0]) << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  fact[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    fact[i] = mult(fact[i - 1], i);
  }
  int t = 1;
  cin >> t;
  while (t--) solve();
}

```