#include <bits/stdc++.h>
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,abm,mmx,avx,avx2,popcnt")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T1, typename T2>
inline void chkmin(T1& x, const T2& y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1& x, const T2& y) {
  if (x < y) x = y;
}
mt19937 rnd(time(0));
int mod = 1000000007;
int mul(int a, int b) { return 1ll * a * b % mod; }
int add(int a, int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
int sub(int a, int b) {
  if (a - b < 0) return a - b + mod;
  return a - b;
}
int binpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
const int MAXN = 2e5 + 228;
int inv(int x) { return binpow(x, mod - 2); }
int fact[MAXN];
int rfact[MAXN];
int cnk(int n, int k) {
  if (k < 0 || k > n) return 0;
  return mul(fact[n], mul(rfact[k], rfact[n - k]));
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout.precision(20), cout.setf(ios::fixed);
  fact[0] = 1;
  for (int i = 1; i < MAXN; ++i) fact[i] = mul(fact[i - 1], i);
  rfact[MAXN - 1] = inv(fact[MAXN - 1]);
  for (int i = MAXN - 2; i >= 0; --i) rfact[i] = mul(rfact[i + 1], i + 1);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    if (n % 2 == 0) {
      int am = 0;
      for (int bt = 0; bt < n; bt += 2) {
        am = add(am, cnk(n, bt));
      }
      int gl = binpow(2, n);
      int ans = 0;
      int nw = 1;
      for (int fbt = k - 1; fbt >= 0; --fbt) {
        ans = add(ans, mul(nw, binpow(gl, fbt)));
        nw = mul(nw, am);
      }
      ans = add(ans, nw);
      cout << ans << endl;
    } else {
      int am = 1;
      for (int bt = 0; bt <= n; bt += 2) {
        am = add(am, cnk(n, bt));
      }
      am = binpow(am, k);
      cout << am << '\n';
    }
  }
}

### 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>
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,abm,mmx,avx,avx2,popcnt")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T1, typename T2>
inline void chkmin(T1& x, const T2& y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1& x, const T2& y) {
  if (x < y) x = y;
}
mt19937 rnd(time(0));
int mod = 1000000007;
int mul(int a, int b) { return 1ll * a * b % mod; }
int add(int a, int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
int sub(int a, int b) {
  if (a - b < 0) return a - b + mod;
  return a - b;
}
int binpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
const int MAXN = 2e5 + 228;
int inv(int x) { return binpow(x, mod - 2); }
int fact[MAXN];
int rfact[MAXN];
int cnk(int n, int k) {
  if (k < 0 || k > n) return 0;
  return mul(fact[n], mul(rfact[k], rfact[n - k]));
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout.precision(20), cout.setf(ios::fixed);
  fact[0] = 1;
  for (int i = 1; i < MAXN; ++i) fact[i] = mul(fact[i - 1], i);
  rfact[MAXN - 1] = inv(fact[MAXN - 1]);
  for (int i = MAXN - 2; i >= 0; --i) rfact[i] = mul(rfact[i + 1], i + 1);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    if (n % 2 == 0) {
      int am = 0;
      for (int bt = 0; bt < n; bt += 2) {
        am = add(am, cnk(n, bt));
      }
      int gl = binpow(2, n);
      int ans = 0;
      int nw = 1;
      for (int fbt = k - 1; fbt >= 0; --fbt) {
        ans = add(ans, mul(nw, binpow(gl, fbt)));
        nw = mul(nw, am);
      }
      ans = add(ans, nw);
      cout << ans << endl;
    } else {
      int am = 1;
      for (int bt = 0; bt <= n; bt += 2) {
        am = add(am, cnk(n, bt));
      }
      am = binpow(am, k);
      cout << am << '\n';
    }
  }
}

```