#include <bits/stdc++.h>
using namespace std;
using li = long long;
using ld = long double;
using pii = pair<int, int>;
const int INF = 1e9 + 13;
const int N = 1e6 + 13;
const int M = 1e9 + 7;
const int B = 600;
const int A = 256;
const ld e = 1e-8;
const int LOGN = 20;
mt19937 rnd(time(0));
int sum(int a, int b) {
  a += b;
  return (a >= M ? a - M : a);
}
int dif(int a, int b) { return sum(a, M - b); }
int mul(int a, int b) { return a * 1ll * b % M; }
int pow2(int n, int k) {
  if (k == 0) return 1;
  int pw = pow2(n, k / 2);
  pw = mul(pw, pw);
  return (k % 2 == 1 ? mul(pw, n) : pw);
}
int pw[N];
void solve() {
  int n, k;
  cin >> n >> k;
  int ans = 0;
  int cur = 1;
  for (int i = k - 1; i >= 0; i--) {
    if (n % 2 == 0) {
      ans = sum(ans, mul(pow2(pw[i], n), cur));
      cur = mul(cur, dif(pw[n - 1], 1));
    } else {
      cur = mul(cur, sum(pw[n - 1], 1));
    }
  }
  cout << sum(ans, cur) << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  pw[0] = 1;
  for (int i = 1; i < N; i++) pw[i] = mul(pw[i - 1], 2);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Create a solution in Cpp for 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;
using li = long long;
using ld = long double;
using pii = pair<int, int>;
const int INF = 1e9 + 13;
const int N = 1e6 + 13;
const int M = 1e9 + 7;
const int B = 600;
const int A = 256;
const ld e = 1e-8;
const int LOGN = 20;
mt19937 rnd(time(0));
int sum(int a, int b) {
  a += b;
  return (a >= M ? a - M : a);
}
int dif(int a, int b) { return sum(a, M - b); }
int mul(int a, int b) { return a * 1ll * b % M; }
int pow2(int n, int k) {
  if (k == 0) return 1;
  int pw = pow2(n, k / 2);
  pw = mul(pw, pw);
  return (k % 2 == 1 ? mul(pw, n) : pw);
}
int pw[N];
void solve() {
  int n, k;
  cin >> n >> k;
  int ans = 0;
  int cur = 1;
  for (int i = k - 1; i >= 0; i--) {
    if (n % 2 == 0) {
      ans = sum(ans, mul(pow2(pw[i], n), cur));
      cur = mul(cur, dif(pw[n - 1], 1));
    } else {
      cur = mul(cur, sum(pw[n - 1], 1));
    }
  }
  cout << sum(ans, cur) << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  pw[0] = 1;
  for (int i = 1; i < N; i++) pw[i] = mul(pw[i - 1], 2);
  int t;
  cin >> t;
  while (t--) solve();
}

```