#include <bits/stdc++.h>
using namespace std;
inline void rd(char &x) { scanf("%c", &x); }
inline void rd(char *s) { scanf("%s", s); }
inline void rd(int &x) { scanf("%d", &x); }
inline void rd(long long &x) { scanf("%lld", &x); }
template <typename T, typename... U>
inline void rd(T &a, U &...b) {
  rd(a), rd(b...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 214514;
const long long mod = 1e9 + 7;
long long pw[maxn] = {1};
long long n, k;
long long qpow(long long a, long long b) {
  b %= (mod - 1);
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ret;
}
long long solve(int k) {
  long long ret = 1;
  for (int i = 0, __i = k - 1; i <= __i; ++i) {
    if (n & 1) {
      ret = (1 + pw[n - 1]) * ret % mod;
    } else {
      ret = qpow(2, i * n) + (pw[n - 1] - 1) * ret % mod;
      ret = (ret + mod) % mod;
    }
  }
  return ret;
}
int main() {
  for (int i = 1, __i = 200100; i <= __i; ++i) pw[i] = pw[i - 1] * 2 % mod;
  int T;
  rd(T);
  for (int kase = 1, __kase = T; kase <= __kase; ++kase) {
    rd(n, k);
    printf("%lld\n", solve(k));
  }
}

### Prompt

In CPP, your task is to solve 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;
inline void rd(char &x) { scanf("%c", &x); }
inline void rd(char *s) { scanf("%s", s); }
inline void rd(int &x) { scanf("%d", &x); }
inline void rd(long long &x) { scanf("%lld", &x); }
template <typename T, typename... U>
inline void rd(T &a, U &...b) {
  rd(a), rd(b...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 214514;
const long long mod = 1e9 + 7;
long long pw[maxn] = {1};
long long n, k;
long long qpow(long long a, long long b) {
  b %= (mod - 1);
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ret;
}
long long solve(int k) {
  long long ret = 1;
  for (int i = 0, __i = k - 1; i <= __i; ++i) {
    if (n & 1) {
      ret = (1 + pw[n - 1]) * ret % mod;
    } else {
      ret = qpow(2, i * n) + (pw[n - 1] - 1) * ret % mod;
      ret = (ret + mod) % mod;
    }
  }
  return ret;
}
int main() {
  for (int i = 1, __i = 200100; i <= __i; ++i) pw[i] = pw[i - 1] * 2 % mod;
  int T;
  rd(T);
  for (int kase = 1, __kase = T; kase <= __kase; ++kase) {
    rd(n, k);
    printf("%lld\n", solve(k));
  }
}

```