#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int M = 2000 + 5;
const int OO = 1e9;
const int mod = 1e9 + 7;
int n, k;
long long fact[N], inv[N], cur, nxt;
long long mul(long long a, long long b) { return (a * b) % mod; }
long long add(long long a, long long b) { return (a + b) % mod; }
long long sub(long long a, long long b) { return (a - b + mod) % mod; }
long long nCr(int n, int r) {
  return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;
}
void calcCurAndNxt() {
  cur = nxt = 0;
  for (int i = 0; i <= n; ++i) {
    bool isOneXOR = (i & 1);
    bool isOneAND = (i == n);
    if (isOneAND == isOneXOR) {
      nxt = add(nxt, nCr(n, i));
    } else if (isOneXOR) {
      cur = add(cur, nCr(n, i));
    }
  }
}
long long pw(long long a, long long b) {
  if (!b) {
    return 1;
  }
  auto ret = pw(a, b / 2);
  ret = mul(ret, ret);
  if (b & 1) {
    ret = mul(ret, a);
  }
  return ret;
}
void calcFacAndInv() {
  fact[0] = inv[0] = 1;
  for (long long i = 1; i < N; i++) {
    fact[i] = mul(i, fact[i - 1]);
    inv[i] = pw(fact[i], mod - 2);
  }
}
long long solve(int bit) {
  if (!bit) {
    return 0;
  }
  return add(mul(cur, pw(2, 1LL * (bit - 1) * n)), mul(nxt, solve(bit - 1)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  calcFacAndInv();
  while (tt--) {
    cin >> n >> k;
    calcCurAndNxt();
    cout << sub(pw(2, 1LL * k * n), solve(k)) << '\n';
  }
  return 0;
}

### Prompt

Generate 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 N = 2e5 + 5;
const int M = 2000 + 5;
const int OO = 1e9;
const int mod = 1e9 + 7;
int n, k;
long long fact[N], inv[N], cur, nxt;
long long mul(long long a, long long b) { return (a * b) % mod; }
long long add(long long a, long long b) { return (a + b) % mod; }
long long sub(long long a, long long b) { return (a - b + mod) % mod; }
long long nCr(int n, int r) {
  return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;
}
void calcCurAndNxt() {
  cur = nxt = 0;
  for (int i = 0; i <= n; ++i) {
    bool isOneXOR = (i & 1);
    bool isOneAND = (i == n);
    if (isOneAND == isOneXOR) {
      nxt = add(nxt, nCr(n, i));
    } else if (isOneXOR) {
      cur = add(cur, nCr(n, i));
    }
  }
}
long long pw(long long a, long long b) {
  if (!b) {
    return 1;
  }
  auto ret = pw(a, b / 2);
  ret = mul(ret, ret);
  if (b & 1) {
    ret = mul(ret, a);
  }
  return ret;
}
void calcFacAndInv() {
  fact[0] = inv[0] = 1;
  for (long long i = 1; i < N; i++) {
    fact[i] = mul(i, fact[i - 1]);
    inv[i] = pw(fact[i], mod - 2);
  }
}
long long solve(int bit) {
  if (!bit) {
    return 0;
  }
  return add(mul(cur, pw(2, 1LL * (bit - 1) * n)), mul(nxt, solve(bit - 1)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  calcFacAndInv();
  while (tt--) {
    cin >> n >> k;
    calcCurAndNxt();
    cout << sub(pw(2, 1LL * k * n), solve(k)) << '\n';
  }
  return 0;
}

```