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 {
      public static void main(String[] args) throws IOException {
          long[] powers = new long[200001];
          powers[0] = 1;
          for (int i = 1; i <= 200000; i++) {
              powers[i] = (powers[i-1]*2l)%1000000007l;
          }
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          int t = Integer.parseInt(br.readLine());
          for (int i = 0; i < t; i++) {
              StringTokenizer st = new StringTokenizer(br.readLine());
              int n = Integer.parseInt(st.nextToken());
              int k = Integer.parseInt(st.nextToken());
              long[] a = new long[k+1];
              long[] powersofpowers = new long[k+1];
              a[0] = 1l;
              powersofpowers[0] = 1l;
              if (n % 2 == 0) {
                  for (int j = 1; j <= k; j++) {
                      powersofpowers[j] = (powersofpowers[j-1]*powers[n])%1000000007l;
                      a[j] = ((powers[n-1]-1)*a[j-1] + powersofpowers[j-1])%1000000007l; //even number of 1's+all solutions for length j-1 
                      //all 1's+all solutions and nonsolutions for length j-1
                  }
              } else {
                  for (int j = 1; j <= k; j++) {
                      a[j] = (powers[n-1]*a[j-1] + a[j-1])%1000000007l; //even number of 1's+all solutions for length j-1 
                      //all 1's+all solutions for length j-1
                  }
              }
              System.out.println(a[k]);
          }
      }
  }