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_Moamen_and_XOR{
      
  
      public static void main(String[] args) {
          FastScanner s= new FastScanner();
          StringBuilder res = new StringBuilder();
          int t=s.nextInt();
          int p=0;
          while(p<t){
          int n=s.nextInt();
          int k=s.nextInt();
          long mod=(long)Math.pow(10,9)+7;
          if(k==0){
              res.append("1\n");
              p++;
              continue;
          }
          if(n%2!=0){
              long yo=power(2,n-1,mod);
              yo++;
              long ans=power(yo,k,mod);
              res.append(ans+"\n");
  
  
          }
          else{
            long ans=0;
            for(int i=k;i>=1;i--){
                long naacho=((long)n*(i-1));
                long yo=power(2,naacho,mod);
                long yoyo=power(2,n-1,mod);
                yoyo--;
                long nice=power(yoyo,k-i,mod);
                long bete=(yo*nice)%mod;
                ans+=bete;
                ans=ans%mod;
            }
            long yo=power(2,n-1,mod);
              yo--;
              long mauj=power(yo,k,mod);
              ans+=mauj;
              ans=ans%mod;
            
            res.append(ans+"\n");
          }
          p++;
      }
      System.out.println(res);
  
      }
      private static long power(long i, long j,long mod) {
          long num=i;
          long ans=1;
          while(j>0){
              long val=j&1;
              if(val==1){
                  ans*=num;
                  ans=ans%mod;
              }
              num=(num*num)%mod;
              j=j>>1;
          }
          return ans;
      }
  
      static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
   
  		public FastScanner(String s) {
  			try {
  				br = new BufferedReader(new FileReader(s));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
  		}
   
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
   
  		String nextToken() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					// TODO Auto-generated catch block
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
   
  		int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
   
  		long nextLong() {
  			return Long.parseLong(nextToken());
  		}
   
  		double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  	}
  
  
  
  
  }