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
  {
  	public static void main(String[] args)throws Exception
  	{
  	    Main ob=new Main();
  	    ob.fun();
  	}
  	public void fun()throws Exception
  	{
  	    
  	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  	    PrintWriter pw=new PrintWriter(System.out);
  	    long mod=1000000007l;
  	    /*long two[]=new long[200009];
  	    long n2=1l;
  	    for(int i=0;i<200009;i++)
  	    {
  	        two[i]=n2%mod;
  	        n2%=mod;
  	        n2*=2l;
  	        n2%=mod;
  	    }*/
  	    int t=Integer.parseInt(br.readLine());
  	    
  	    while(t-->0)
  	    {
  	        String s1[]=(br.readLine()).split(" ");
  	        int n=Integer.parseInt(s1[0]);
  	        int k=Integer.parseInt(s1[1]);
  	        
  	        long even=power2(2l,(long)n-1,mod);
  	        
  	        if(n%2==0)
  	        {
  	            even--;
  	            even+=mod;
  	            even%=mod;
  	        }
  	        else
  	        {
  	        	even++;
  	        	even+=mod;
  	            even%=mod;
  	            long ans2=1l;
  	            for(int i=1;i<=k;i++)
  	            {
  	            	ans2=((ans2%mod)*(even%mod))%mod;
  	            }
  	            pw.println(ans2);
  	            continue;
  	        }
  	        long mulE=1l;
  	        long ans=0l;
  	        for(int i=0;i<k;i++)
  	        {
  	            long prod=mulE;
  	            long power=n;
  	            power*=(long)(k-1-i);
  	            long prod2=power2(2l,power,mod);
  	            prod2%=mod;
  	            prod=((prod%mod)*(prod2%mod))%mod;
  	            ans=(ans%mod + prod%mod)%mod;
  	            mulE=((mulE%mod)*(even%mod))%mod;
  	        }
  	        
  	        ans=((ans%mod)+(mulE%mod))%mod;
  	        pw.println(ans);
  	    }
  	        
  	    
  	    pw.flush();
  	}
  	public long power2(long base,long p,long mod)
  	{
  	    if(p==0)
  	    {
  	        return 1l;
  	    }
  	    if(p%2==0)
  	    {
  	        return power2((base*base)%mod,p/2,mod);
  	    }
  	    else
  	    {
  	        return (((base%mod)*(power2((base*base)%mod,p/2,mod))%mod)%mod);
  	    }
  	}
  
  }