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.Arrays;
  import java.util.Scanner;
  
  public class C {
      public static void main(String[] args) {
          Scanner s = new Scanner(System.in);
          int lines = s.nextInt();
  
          s.nextLine();
  
          for (int i = 0; i < lines; i += 1) {
              solve(s.nextInt(), s.nextInt());
          }
      }
  
      public static void solve(int l, int k) {
          pow = new long[l + 1];
          pow[0] = 1;
          for (int i = 1; i <= l; i++) {
              pow[i] = (pow[i - 1] * 2) % mod;
          }
  
          sol = new long[k + 1];
          sol[0] = 1;
          for (int i = 1; i <= k; i++) {
              sol[i] = rec(l, i, sol[i - 1] % mod);
          }
  
          //System.out.println(Arrays.toString(sol));
          System.out.println(sol[k] % mod);
      }
  
      static long sol[];
      static long pow[];
      static long mod = 1000000007L;
  
      public static long rec(int l, long k, long last) {
          if (k == 0) {
              return 1;
          }
  
          long erg;
          if (l % 2 == 0) {
              erg = (((pow[l - 1] - 1) * last) % mod + pow(pow[l], k - 1)) % mod;
          } else {
              erg = ((pow[l - 1] + 1) * last) % mod;
          }
          return erg;
      }
  
      public static long pow(long b, long e) {
          if (e == 0) return 1;
          long l = pow(b, e / 2);
          return e % 2 == 0 ? (l * l) % mod : (((l * l) % mod) * b) % mod;
      }
  }
  