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 {
      private static long[] factorial;
      private static final int MOD = 1000000007;
      private static int multiply(long a, long b) {
          return (int) ((a%MOD*b%MOD)%MOD);
      }
      private static int divide(long a, long b) {
          return multiply(a, modularInverse(b));
      }
      private static int modularInverse(long a) {
          return power(a, MOD-2);
      }
      private static int power(long a, long b) {
          long c = 1;
          while(b > 0) {
              if((b&1) > 0) {
                  c = multiply(c, a);
              }
              a = multiply(a, a);
              b >>= 1;
          }
          return (int) c;
      }
      private static void calculateFactorials() {
          factorial[0] = 1;
          for(int i = 1; i < factorial.length; i++) {
              factorial[i] = multiply(factorial[i-1], i);
          }
      }
      private static int C(int n, int r) {
          if(r > n) {
              return 0;
          }
          return divide(factorial[n], multiply(factorial[r], factorial[n-r]));
      }
      public static void main(String[] args) throws IOException {
          //BufferedReader f = new BufferedReader(new FileReader("uva.in"));
          BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
          factorial = new long[200001];
          calculateFactorials();
          int t = Integer.parseInt(f.readLine());
          while(t-- > 0) {
              StringTokenizer st = new StringTokenizer(f.readLine());
              int n = Integer.parseInt(st.nextToken());
              int k = Integer.parseInt(st.nextToken());
              long sum = 0;
              for(int i = 2; i < n; i += 2) {
                  sum = (sum+C(n, i))%MOD;
              }
              long temp = 1;
              for(int i = 0; i < n; i++) {
                  temp = multiply(temp, 2);
              }
              long ans = 1;
              for(int i = 1; i <= k; i++) {
                  long cur;
                  if(n%2 == 0) {
                      cur = multiply(sum, ans);
                      cur = (cur+power(temp, i-1))%MOD;
                  } else {
                      cur = multiply(sum+1, ans);
                  }
                  ans = (ans+cur)%MOD;
              }
              out.println(ans);
          }
          f.close();
          out.close();
      }
  }
  