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.io.PrintWriter;
  import java.util.Arrays;
  
  public class Practice {
      public static long find(long n, long k) {
          if (k == 0) return 1;
          long mod = 1000000007L;
          boolean even = (n % 2 == 0);
          long c1 = even ? 1: 0;
          long c2 = even ? 0: 1;
          long c3 = even ? up(n - 1, mod) - 1: up(n - 1, mod);
          long last = 1;
          for (int i = 1; i <= k; i++) {
              long a = c1 * up((i - 1) * n, mod);
              long b = c2 * last;
              long c = c3 * last;
              long sum = ((a % mod) + (b % mod)) % mod;
              sum = (sum + (c % mod)) % mod;
              last = sum;
          }
          return last;
      }
      
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter pw = new PrintWriter(System.out);
          int t = Integer.parseInt(br.readLine());
          for (int j = 0; j < t; j++) {
              String[] in = br.readLine().split("\\s");
              long n = Integer.parseInt(in[0]);
              long k = Integer.parseInt(in[1]);
              pw.println(find(n, k));
          }
          pw.flush();
      }
  
      private static long up(long power, long mod) {
          if (power == 0) return 1;
          long x = up(power / 2, mod);
          x = (x * x) % mod;
          if (power % 2 == 1) x *= 2;
          return x % mod;
      }
  
      /*public static void main(String[] args) {
          long mod = 1000000007;
          System.out.println(up(3, mod));
          System.out.println(up(4, mod));
          System.out.println(up(5, mod));
          System.out.println(up(100, mod));
      }*/
  }