#include <bits/stdc++.h>
using namespace std;
void run() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const int N = 2e5 + 1;
const int MOD = 1e9 + 7;
long long fact[N], inv[N], invfact[N];
void factInverse() {
  fact[0] = inv[1] = fact[1] = invfact[0] = invfact[1] = 1;
  for (long long i = 2; i < N; i++) {
    fact[i] = (fact[i - 1] * i) % MOD;
    inv[i] = MOD - (inv[MOD % i] * (MOD / i) % MOD);
    invfact[i] = (inv[i] * invfact[i - 1]) % MOD;
  }
}
int add(int a, int b) {
  if ((a += b) >= MOD)
    a -= MOD;
  else if (a < 0)
    a += MOD;
  return a;
}
long long mul(int x, int y) { return (1LL * x * y) % MOD; }
long long nCr(int n, int r) {
  if (r > n) return 0;
  return mul(mul(fact[n], invfact[r]), invfact[n - r]);
}
int mem[N][2], vis[N][2], test_id;
int cntEven, _2pwn, n;
int dp(int i, bool equal) {
  if (i < 0) return 1;
  int& rt = mem[i][equal];
  if (vis[i][equal] == test_id) return rt;
  vis[i][equal] = test_id;
  if (equal == false) return rt = mul(_2pwn, dp(i - 1, 0));
  if (n & 1) return rt = mul(dp(i - 1, 1), add(cntEven, 1));
  return rt = add(mul(dp(i - 1, 1), cntEven), dp(i - 1, 0));
}
int main() {
  run();
  factInverse();
  int t;
  cin >> t;
  while (t--) {
    test_id++;
    int k;
    cin >> n >> k;
    cntEven = 0;
    for (int i = 0; i < n; i += 2) cntEven = add(cntEven, nCr(n, i));
    _2pwn = 1;
    for (int i = 0; i < n; i++) _2pwn = mul(_2pwn, 2);
    cout << dp(k - 1, 1) << "\n";
  }
}

### Prompt

Please create a solution in Cpp 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;
void run() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const int N = 2e5 + 1;
const int MOD = 1e9 + 7;
long long fact[N], inv[N], invfact[N];
void factInverse() {
  fact[0] = inv[1] = fact[1] = invfact[0] = invfact[1] = 1;
  for (long long i = 2; i < N; i++) {
    fact[i] = (fact[i - 1] * i) % MOD;
    inv[i] = MOD - (inv[MOD % i] * (MOD / i) % MOD);
    invfact[i] = (inv[i] * invfact[i - 1]) % MOD;
  }
}
int add(int a, int b) {
  if ((a += b) >= MOD)
    a -= MOD;
  else if (a < 0)
    a += MOD;
  return a;
}
long long mul(int x, int y) { return (1LL * x * y) % MOD; }
long long nCr(int n, int r) {
  if (r > n) return 0;
  return mul(mul(fact[n], invfact[r]), invfact[n - r]);
}
int mem[N][2], vis[N][2], test_id;
int cntEven, _2pwn, n;
int dp(int i, bool equal) {
  if (i < 0) return 1;
  int& rt = mem[i][equal];
  if (vis[i][equal] == test_id) return rt;
  vis[i][equal] = test_id;
  if (equal == false) return rt = mul(_2pwn, dp(i - 1, 0));
  if (n & 1) return rt = mul(dp(i - 1, 1), add(cntEven, 1));
  return rt = add(mul(dp(i - 1, 1), cntEven), dp(i - 1, 0));
}
int main() {
  run();
  factInverse();
  int t;
  cin >> t;
  while (t--) {
    test_id++;
    int k;
    cin >> n >> k;
    cntEven = 0;
    for (int i = 0; i < n; i += 2) cntEven = add(cntEven, nCr(n, i));
    _2pwn = 1;
    for (int i = 0; i < n; i++) _2pwn = mul(_2pwn, 2);
    cout << dp(k - 1, 1) << "\n";
  }
}

```