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 CodeForces
  {
  	static FastReader sc=new FastReader(System.in);
  	static  PrintWriter out=new PrintWriter(System.out);
  	static class FastReader
  	{
          byte[] buf = new byte[2048];
          int index, total;
          InputStream in;
   
          FastReader(InputStream is) 
          {
              in = is;
          }
   
          int scan() throws IOException 
          {
              if (index >= total) {
                  index = 0;
                  total = in.read(buf);
                  if (total <= 0) {
                      return -1;
                  }
              }
              return buf[index++];
          }
   
          String next() throws IOException 
          {
              int c;
              for (c = scan(); c <= 32; c = scan());
              StringBuilder sb = new StringBuilder();
              for (; c > 32; c = scan()) {
                  sb.append((char) c);
              }
              return sb.toString();
          }
   
          int nextInt() throws IOException 
          {
              int c, val = 0;
              for (c = scan(); c <= 32; c = scan());
              boolean neg = c == '-';
              if (c == '-' || c == '+') {
                  c = scan();
              }
              for (; c >= '0' && c <= '9'; c = scan()) {
                  val = (val << 3) + (val << 1) + (c & 15);
              }
              return neg ? -val : val;
          }
   
          long nextLong() throws IOException 
          {
              int c;
              long val = 0;
              for (c = scan(); c <= 32; c = scan());
              boolean neg = c == '-';
              if (c == '-' || c == '+') {
                  c = scan();
              }
              for (; c >= '0' && c <= '9'; c = scan()) {
                  val = (val << 3) + (val << 1) + (c & 15);
              }
              return neg ? -val : val;
          }
      }
  	public static void main(String[] args) throws java.lang.Exception
  	{
  		int T=1;
  		T=sc.nextInt();
  		StringBuilder sb=new StringBuilder();
  		while(T-->0)
  		{
  			long n=sc.nextLong(),k=sc.nextLong();
  			if((n&1)!=0)
  				sb.append(power(power(2,n-1)+1,k));
  			else
  			{
  				long ans=power(power(2,n-1)-1,k);
  				for(long i=k;i>=1;i--)
  				{
  					ans+=((power(power(2,n),i-1)%1000000007)*(power(power(2,n-1)-1,k-i)%1000000007))%1000000007;
  					ans%=1000000007;
  				}
  				sb.append(ans);
  			}
  			sb.append('\n');
  		}
  		out.print(sb);
  		out.flush();
  		out.close();
  	}
  //	static long power(long x, long y)
  //	{
  //	    long temp;
  //	    if( y == 0)
  //	        return 1;
  //	    temp = power(x, y / 2)%1000000007;
  //	    if (y % 2 == 0)
  //	        return (temp*temp)%1000000007;
  //	    else 
  //	        return (x*(temp*temp)%1000000007)%1000000007;
  //	}
  	static long power(long x, long y)
  	  {
  		long p=1000000007;
  	    long res = 1; // Initialize result
  	 
  	    x = x % p; // Update x if it is more than or
  	    // equal to p
  	 
  	    if (x == 0)
  	      return 0; // In case x is divisible by p;
  	 
  	    while (y > 0)
  	    {
  	 
  	      // If y is odd, multiply x with result
  	      if ((y & 1) != 0)
  	        res = (res * x) % p;
  	 
  	      // y must be even now
  	      y = y >> 1; // y = y/2
  	      x = (x * x) % p;
  	    }
  	    return res%p;
  	  }
  }
  
  
  
  
  
  
  
  
  
  
  