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.*;
  import java.util.*;
  
  public class Main {
  
      static final int MOD = 1000000007;
  
      public static void main(String[] args) {
  
          MyScanner scan = new MyScanner();
          out = new PrintWriter(new BufferedOutputStream(System.out));
  
          for (int i = scan.nextInt(); i > 0; i--) {
              int n = scan.nextInt();
              int k = scan.nextInt();
  
              int x = 1;
              for (int j = 1; j <= k ; j++) {
                  x = calc(n, j, x);
              }
              out.println(x);
          }
          out.close();
      }
  
      static int calc(int n, int k, long poss){
          poss *= potMod(n-1) - 1 + 2*(n % 2);
          if(n % 2 == 0)poss += potMod((int) ((long)(k-1)* n % (MOD-1)));
          poss %= MOD;
          return (int) poss;
      }
  
      static int potMod(int exp){
          long pot = 1;
          //exp %= (MOD-1);
          if(exp > 0){
              int temp = potMod(exp/2);
              pot = (long) temp * temp;
              pot *= (1 + exp % 2);
              pot %= MOD;
              return (int) pot;
          }else {
              return 1;
          }
      }
  
  
      public static PrintWriter out;
      public static class MyScanner { BufferedReader br; StringTokenizer st;
  
          public MyScanner() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine(){
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }