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 Codeforces
  {
      public static void main(String args[])throws Exception
      {
          BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
          StringBuilder sb=new StringBuilder();
          int t=Integer.parseInt(bu.readLine());
          int N=200000,i;
          long p[]=new long[N+1],M=(int)1e9+7;
          p[0]=1;
          for(i=1;i<=N;i++) p[i]=p[i-1]*2%M;
  
          while(t-->0)
          {
              String s[]=bu.readLine().split(" ");
              int n=Integer.parseInt(s[0]),k=Integer.parseInt(s[1]);
              if(k==0) {sb.append("1\n"); continue;}
              long dp[]=new long[k];
              for(i=0;i<k;i++)
              {
                  long win=0,draw=p[n-1];
                  if(n%2==0) {win++; draw--;}
                  else draw++;
                  if(i==0) dp[i]=win+draw;
                  else dp[i]=(win*power(p[n],i,M)%M+dp[i-1]*draw%M)%M;
              }
              sb.append(dp[k-1]+"\n");
          }
          System.out.print(sb);
      }
  
      static long power(long a,int b,long M)
      {
          long res=1;
          while(b>0)
          {
              if(b%2==1) res=res*a%M;
              b>>=1;
              a=a*a%M;
          }
          return res;
      }
  }
  