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.*;
  import java.math.*;
  
  public class Main
  {
  	public static void main(String[] args)throws IOException
  	{
  		BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
  		PrintWriter out = new PrintWriter(System.out);
  		BigInteger MOD = new BigInteger("1000000007");
  		int t = Integer.parseInt(input.readLine());
  		while(t-->0)
  		{
  			StringTokenizer st = new StringTokenizer(input.readLine());
  			int n = Integer.parseInt(st.nextToken());
  			int k = Integer.parseInt(st.nextToken());
  			BigInteger bi = new BigInteger("1");
  			if(n%2==1)
  			{
  				int temp = n/2+1;
  				BigInteger four = new BigInteger("4");
  				BigInteger exp = new BigInteger(temp-1+"");
  				BigInteger factor = four.modPow(exp,MOD);
  				factor = factor.add(BigInteger.ONE).mod(MOD); 
  				for(int i = 1; i <= k; i++)
  				{
  					bi = bi.multiply(factor).mod(MOD);
  				}
  			}
  			else
  			{
  				int temp = n/2-1;
  				BigInteger four = new BigInteger("4");
  				BigInteger exp = new BigInteger(temp+"");
  				BigInteger factor = four.modPow(exp,MOD);
  				factor = factor.multiply(new BigInteger("2")).add(MOD).subtract(BigInteger.ONE).mod(MOD);
  				BigInteger addition = BigInteger.ONE;
  				BigInteger exp2 = new BigInteger(n/2+"");
  				BigInteger additionFactor = four.modPow(exp2,MOD); 
  				for(int i = 1; i <= k; i++)
  				{
  					bi = bi.multiply(factor).add(addition).mod(MOD);
  					addition = addition.multiply(additionFactor).mod(MOD);
  				}
  			}
  			out.println(bi.add(MOD).mod(MOD));
  		}
  		out.flush();
  		out.close();
  	}
  }