#include <iostream>

const int MOD = 1000000007;

using namespace std;
using ll = long long;

ll pow(ll b, ll e) {
    if (e == 0) return 1;
    if (e == 1) return b;
    ll prev = pow(b, e / 2);
    return (((prev * prev) % MOD) * ((e % 2 == 1) ? (b) : (1))) % MOD;
}

ll solve(int n, int k, const ll& win, const ll& tie) {
    if (k == 0) return 1;
    // win
    ll w = (win * pow(pow(2, n), k - 1)) % MOD;
    // win later
    ll wl = (tie * solve(n, k - 1, win, tie)) % MOD;
    return (w + wl) % MOD;
}

void solve() {
    int n, k;
    cin >> n >> k;
    if (k == 0) {
        cout << 1 << endl;
        return;
    }
    ll kk = pow(2, n - 1);
    ll win, tie;
    if (n % 2 == 0) {
        win = 1;
        tie = kk - 1;
    } else if (n % 2 == 1) {
        win = 0;
        tie = kk + 1;
    }
    ll r = solve(n, k, win, tie);
    cout << r << endl;
}

int main() {
    int t;
    cin >> t;
    while (t--)
        solve();

    return 0;
}
### 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 <iostream>

const int MOD = 1000000007;

using namespace std;
using ll = long long;

ll pow(ll b, ll e) {
    if (e == 0) return 1;
    if (e == 1) return b;
    ll prev = pow(b, e / 2);
    return (((prev * prev) % MOD) * ((e % 2 == 1) ? (b) : (1))) % MOD;
}

ll solve(int n, int k, const ll& win, const ll& tie) {
    if (k == 0) return 1;
    // win
    ll w = (win * pow(pow(2, n), k - 1)) % MOD;
    // win later
    ll wl = (tie * solve(n, k - 1, win, tie)) % MOD;
    return (w + wl) % MOD;
}

void solve() {
    int n, k;
    cin >> n >> k;
    if (k == 0) {
        cout << 1 << endl;
        return;
    }
    ll kk = pow(2, n - 1);
    ll win, tie;
    if (n % 2 == 0) {
        win = 1;
        tie = kk - 1;
    } else if (n % 2 == 1) {
        win = 0;
        tie = kk + 1;
    }
    ll r = solve(n, k, win, tie);
    cout << r << endl;
}

int main() {
    int t;
    cin >> t;
    while (t--)
        solve();

    return 0;
}
```