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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  
  public class MM {
  	static long mod = 1000000007;
      public static void main(String[] arg){
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
          while(t-->0){
              int n=sc.nextInt();
              int k=sc.nextInt();
              if(k==0){
                  System.out.println(1);
                  continue;
              }
              if(n%2==1){
                  long base = pow(2, n-1) + 1;
                  long ans = pow(base%mod, k);
                  System.out.println(ans);
                  continue;
              }
   
              long b = pow(2, n-1);
              System.out.println(compute2(n, k, b));
          }
      }
   
      static long compute2(long n, long k, long b){
          if(k==0)return 1;
          long a = compute2(n, k-1, b);
          long ans = pow(b*2%mod, k-1);
          a = (a * (b-1)) % mod;
          ans = (ans + a) % mod;
          return ans;
      }
   
      static long pow(long a, long b){//快速幂
          if(b==0) return 1;
          long ans = b%2==1?a : 1;
          if(b==1)return ans%mod;
          long p = pow(a, b/2);
          ans = (ans * p) % mod;
          ans = (ans * p) % mod;
          return ans;
      }
  }
  
  