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 {
  	static long mod=1000000007;
  	
  	static long mpow(long a, long b) {
  		long res=1;
  		a%=mod;
  		while(b !=0) {
  			if(b%2==1) {
  				res*=a;
  			}
  			b/=2;
  			a*=a;
  			a%=mod;
  			res%=mod;
  		}
  		return res;
  	}
  		
  	public static void main(String[] args) throws IOException {
  		// TODO Auto-generated method stub
  		Reader.init(System.in);
  		//Reader.init();
  		PrintWriter out=new PrintWriter(System.out);
  		long[] pow=new long[200001];
  		pow[0]=1;
  		for(int i=1;i<=200000;i++) {
  			pow[i]= (2*pow[i-1])%mod;
  		}
  		int t=Reader.nextInt();
  		while(t>0) {
  			t--;
  			int n=Reader.nextInt(),  k=Reader.nextInt();
  			if(k==0) {
  				out.println(1);continue;
  			}
  			if(n%2==1) {
  				out.println((mpow((pow[n-1]+1),k)));
  			}
  			else {
  				long[] dp=new long[k+1];
  				dp[1]= pow[n-1];
  				dp[1]%=mod;
  				for(int i=2;i<=k;i++) {
  					dp[i]=mpow(pow[n],i-1);
  					long temp= pow[n-1]-1+mod;
  					temp%=mod;
  					temp*= dp[i-1];
  					temp%=mod;
  					dp[i] += temp;
  					dp[i]%=mod;
  				}
  				out.println(dp[k]);
  			}
  			
  			
  			
  		}
  		
  		
  		
  		out.flush();
  		
  	}
  
  }
  
  
   
  
  
  
  
  
  
  
  
  class Reader {
      static BufferedReader reader;
      static StringTokenizer tokenizer;
      /** call this method to initialize reader for InputStream */
      static void init() throws IOException {
          reader = new BufferedReader(
                  new FileReader("C:\\Users\\samee\\Desktop\\testcase33.txt"));
          tokenizer = new StringTokenizer("");
      }
      static void init(InputStream input) {
          reader = new BufferedReader(
                  new InputStreamReader(input) );
          tokenizer = new StringTokenizer("");
      }
      /** get next word */
      static String nextLine() throws IOException{
          return reader.readLine();
      }
      static String next() throws IOException {
          while ( ! tokenizer.hasMoreTokens() ) {
              //TODO add check for eof if necessary
              tokenizer = new StringTokenizer(
                      reader.readLine() );
          }
          return tokenizer.nextToken();
      }
      static int nextInt() throws IOException {
          return Integer.parseInt( next() );
      }
      static long nextLong() throws IOException {
          return Long.parseLong( next() );
      }
      static double nextDouble() throws IOException {
          return Double.parseDouble( next() );
      }
  }