#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
inline int inc(int x, int y) { return x += y - mod, x += x >> 31 & mod; }
inline int dec(int x, int y) { return x -= y, x += x >> 31 & mod; }
inline int mul(int x, int y) { return long long(x) * y % mod; }
inline int sed(int x, int y) { return y ? x & 1 ? mod - y : y : 0; }
int qpow(int a, int b) {
  b += b >> 31 & (mod - 1);
  int c = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) c = mul(a, c);
  }
  return c;
}
const int maxn = 2e5 + 5;
int T, n, k;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &k);
    int a = qpow(2, n);
    int A = qpow(a, k);
    int I = qpow(a, mod - 2);
    int c = 1;
    int s = 0;
    for (int i = int(0); i <= int(k - 1); i++) {
      int p = dec((mod + 1) / 2, (n & 1 ? I : 0));
      int q = n & 1 ? 0 : I;
      s = inc(s, mul(c, p));
      c = mul(c, dec(1, inc(p, q)));
    }
    printf("%d\n", mul(dec(1, s), A));
  }
  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 mod = 1e9 + 7;
inline int inc(int x, int y) { return x += y - mod, x += x >> 31 & mod; }
inline int dec(int x, int y) { return x -= y, x += x >> 31 & mod; }
inline int mul(int x, int y) { return long long(x) * y % mod; }
inline int sed(int x, int y) { return y ? x & 1 ? mod - y : y : 0; }
int qpow(int a, int b) {
  b += b >> 31 & (mod - 1);
  int c = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) c = mul(a, c);
  }
  return c;
}
const int maxn = 2e5 + 5;
int T, n, k;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &k);
    int a = qpow(2, n);
    int A = qpow(a, k);
    int I = qpow(a, mod - 2);
    int c = 1;
    int s = 0;
    for (int i = int(0); i <= int(k - 1); i++) {
      int p = dec((mod + 1) / 2, (n & 1 ? I : 0));
      int q = n & 1 ? 0 : I;
      s = inc(s, mul(c, p));
      c = mul(c, dec(1, inc(p, q)));
    }
    printf("%d\n", mul(dec(1, s), A));
  }
  return 0;
}

```