#include <bits/stdc++.h>
namespace FastRead {
char __buff[5000];
long long __lg = 0, __p = 0;
char nc() {
  if (__lg == __p) {
    __lg = fread(__buff, 1, 5000, stdin);
    __p = 0;
    if (!__lg) return EOF;
  }
  return __buff[__p++];
}
template <class T>
void read(T& __x) {
  T __sgn = 1;
  char __c;
  while (!isdigit(__c = nc()))
    if (__c == '-') __sgn = -1;
  __x = __c - '0';
  while (isdigit(__c = nc())) __x = __x * 10 + __c - '0';
  __x *= __sgn;
}
}  // namespace FastRead
using namespace FastRead;
using namespace std;
const long long N = 1e5 + 10;
const long long M = 1e9 + 7;
const long double PI = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long n, k;
long long pw(long long n, long long p) {
  long long ans = 1;
  for (; p; p >>= 1, n = n * n % M)
    if (p & 1) ans = ans * n % M;
  return ans;
}
void Test() {
  cin >> n >> k;
  if (n & 1) return cout << pw((pw(2, n - 1) + 1), k) << '\n', void();
  long long X = (pw(2, n - 1) - 1 + M) % M;
  long long ans = pw(X, k);
  for (long long c = 1; c <= k; c++)
    ans = (ans + pw(X, c - 1) * pw(pw(2, n), k - c)) % M;
  cout << ans << '\n';
}
signed main() {
  long long q;
  cin >> q;
  while (q--) Test();
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
namespace FastRead {
char __buff[5000];
long long __lg = 0, __p = 0;
char nc() {
  if (__lg == __p) {
    __lg = fread(__buff, 1, 5000, stdin);
    __p = 0;
    if (!__lg) return EOF;
  }
  return __buff[__p++];
}
template <class T>
void read(T& __x) {
  T __sgn = 1;
  char __c;
  while (!isdigit(__c = nc()))
    if (__c == '-') __sgn = -1;
  __x = __c - '0';
  while (isdigit(__c = nc())) __x = __x * 10 + __c - '0';
  __x *= __sgn;
}
}  // namespace FastRead
using namespace FastRead;
using namespace std;
const long long N = 1e5 + 10;
const long long M = 1e9 + 7;
const long double PI = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long n, k;
long long pw(long long n, long long p) {
  long long ans = 1;
  for (; p; p >>= 1, n = n * n % M)
    if (p & 1) ans = ans * n % M;
  return ans;
}
void Test() {
  cin >> n >> k;
  if (n & 1) return cout << pw((pw(2, n - 1) + 1), k) << '\n', void();
  long long X = (pw(2, n - 1) - 1 + M) % M;
  long long ans = pw(X, k);
  for (long long c = 1; c <= k; c++)
    ans = (ans + pw(X, c - 1) * pw(pw(2, n), k - c)) % M;
  cout << ans << '\n';
}
signed main() {
  long long q;
  cin >> q;
  while (q--) Test();
  return 0;
}

```