description: 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].
solution: import java.util.*;
  import java.io.*;
  
  public class C{
  	private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	private static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
  	private static long mod = 1000000007;
  
  	public static void main (String[] args) throws IOException {
          int t = Integer.parseInt(br.readLine());
          while(t-- > 0) {
              String ip[] = br.readLine().split(" ");
              long n = Long.parseLong(ip[0]);
              long k = Long.parseLong(ip[1]);
              long ncr = power(2, n-1, mod)%mod;
              if(n%2 != 0) {
                  ncr = (ncr + 1)%mod;
                  long ans = power(ncr, k, mod);
                  bw.write(ans + "\n");
              }else {
                  long mult = (ncr - 1)%mod;
                  long left = 1;
                  long any = power(2, n, mod);
                  long ans = power(mult, k, mod);
                  for(int a=0;a<k;a++) {
                      long right = power(any, k-a-1, mod);
                      long cur = (left % mod) * (right % mod) % mod;
                      ans = (ans%mod + cur%mod) % mod;
                      left = (left%mod * mult%mod) % mod;
                  }
                  bw.write(ans + "\n");
              }
              
          }
          bw.flush();
      }
  
      public static long power(long x, long y, long p) {
          long res = 1;
          x = x % p;
          if (x == 0)
              return 0;
      
          while (y > 0) {
              if ((y & 1) != 0)
                  res = (res * x) % p;
              y = y >> 1;
              x = (x * x) % p;
          }
          return res;
      }
  
  }