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: 
  
  pri=pow(10,9)+7
  import sys
  input=sys.stdin.readline
  
  a=int(input())
  for i in range(a):
      n,k=map(int,input().split())
  
      g1=pow(2,n,pri)
      g2=pow(2,n,pri)
     
      if(k==0):
          print(1)
          continue;
  
      ans=[0 for i in range(k)]
  
  
      if(n%2==0):
          ways=pow(2,n-1,pri)
      else:
          ways=pow(2,n-1,pri)
          ways+=1
      ans[0]=ways
      
      for i in range(1,k):
          if(n%2==0):
              ways=pow(2,n-1,pri)
          else:
              ways=pow(2,n-1,pri)
  
              ways+=1
          ways%=pri
  
          if(n%2==0):
              ans[i]=(ans[i-1]*(ways-1))%pri
              ans[i]+=g2
              ans[i]%=pri
              g2=g2*g1
              g2%=pri
              
          else:
              
              ans[i]=ans[i-1]*ways
              ans[i]%=pri
      print(ans[-1])
  
  
  
  
  
              
  
          
  