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.*;
  
  public class Div21557C {
  	
  	static long MOD = 1000000007;
  	static long[] factorial = new long[200001];
  	
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  
  		Scanner s = new Scanner(System.in);
  		
  		factorial[0] = 1;
  		for (int i = 1; i <= 200000; i++) {
  		    factorial[i] = factorial[i - 1] * i % MOD;
  		}
  		
  		int t = s.nextInt();
  		
  		while(t --> 0) {
  			
  			int n = s.nextInt();
  			int k = s.nextInt();
  			
  			//Prep full wipes
  			long prefN = binpow(2, n, MOD);			
  			long[] prefs = new long[k+1];
  			prefs[0] = 1;
  			
  			for(int i = 1; i <= k; i++)
  				prefs[i] = (prefN * prefs[i-1]) % MOD;
  							
  			long bin = 0;
  			
  			for(int i = 0; i < n; i += 2) 
  				bin = (binomial_coefficient(n, i) + bin) % MOD;
  						
  			long[] dp = new long[k+1];
  			dp[0] = 1;
  			
  			for(int i = 1; i <= k; i++) {
  				
  				if(n % 2 == 0)
  					dp[i] = (prefs[i-1] + (dp[i-1] * bin) % MOD) % MOD;
  				else
  					dp[i] = (dp[i-1] * (bin + 1)) % MOD;
  				
  			}
  			
  			System.out.println(dp[k]);			
  		}		
  		
  	}
  	
  	static long binpow(long a, long b, long m) {
  	    a %= m;
  	    long res = 1;
  	    while (b > 0) {
  	        if ((b & 1) == 1)
  	            res = res * a % m;
  	        a = a * a % m;
  	        b >>= 1;
  	    }
  	    return res;
  	}
  	
  	static long binomial_coefficient(int n, int k) {
  		if(k == 0)
  			return 1;
  	    return factorial[n] * binpow(factorial[k] * factorial[n - k] % MOD, MOD-2, MOD) % MOD;
  	}
  
  }
  