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 CF1557C extends PrintWriter {
  	CF1557C() { super(System.out); }
  	Scanner sc = new Scanner(System.in);
  	public static void main(String[] $) {
  		CF1557C o = new CF1557C(); o.main(); o.flush();
  	}
  
  	static final int MD = 1000000007, N = 200000, K = 200000;
  	void main() {
  		int[] pp = new int[N + 1];
  		for (int p = 1, n = 0; n <= N; n++) {
  			pp[n] = p; p = p * 2 % MD;
  		}
  		int[] qq = new int[K];
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int k = sc.nextInt();
  			long gt, eq;
  			if (n % 2 == 0) {
  				gt = 1;
  				eq = pp[n - 1] - 1;
  			} else {
  				gt = 0;
  				eq = pp[n - 1] + 1;
  			}
  			for (int p = pp[n], q = 1, h = 0; h < k; h++) {
  				qq[h] = q; q = (int) ((long) q * p % MD);
  			}
  			long ans = 0, e = 1;
  			for (int h = 0; h < k; h++) {
  				ans = (ans + e * gt % MD * qq[k - 1 - h]) % MD;
  				e = e * eq % MD;
  			}
  			ans = (ans + e) % MD;
  			println(ans);
  		}
  	}
  }
  