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.*;
  import java.io.*;
  
  public class C {
      public static long MOD = 1000000007;
      public static int MAXN = 200005;
      public static void main(String[] args) throws IOException{
          BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
          long[] pow2 = new long[MAXN];
          pow2[0] = 1;
          for(int i = 1; i < MAXN; i++){
              pow2[i] = (pow2[i-1]*2)%MOD;
          }
          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());
              if(k == 0){
                  out.println(1);
                  continue;
              }
              long left = 1;
              long num = ((pow2[n-1]))%MOD;
              long ans = 0;
              for(int temp = k; temp >= 1; temp--){
                  if(temp == 1){
                      if(n % 2 == 0){
                          ans+=((left*((num-1)%MOD))%MOD);
                          ans%=MOD;
                      }else{
                          ans+=((left*((num+1)%MOD))%MOD);
                          ans%=MOD;
                      }
                      break;
                  }
                  if(n % 2 == 0){
                      left*=(num-1);
                      left%=MOD;
                  }else{
                      left*=(num+1);
                      left%=MOD;
                  }
  
              }
              if(n % 2 == 0){
                  for(int i = k; i >= 1; i--){
                      ans+=(pow((num-1), i-1, MOD)*(pow(pow2[n], k-i, MOD))%MOD);
                      ans%=MOD;
                  }
              }
              out.println(ans);
          }
          out.close();
      }
      public static long pow(long a, long b, long p){
          long res = 1;
          while(b > 0){
              if((b & 1) == 1){
                  res*=a;
                  res%=p;
              }
              a = (a*a)%p;
              b/=2;
          }
          return res;
      }
  }
  