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: from __future__ import division,print_function
  from heapq import*
  import sys
  le = sys.__stdin__.read().split("\n")[::-1]
  af=[]
  mo=10**9+7
  def brute(n,k):
      l=[]
      def gen(n,k,ind=0):
          if ind!=n:
              s=0
              for i in range(2**k):
                  l.append(i)
                  s+=gen(n,k,ind+1)
                  l.pop()
              return s
          else:
              an=l[0]
              xo=0
              for i in l:
                  an &= i
                  xo ^= i
              return an>=xo
      return gen(n,k)
  for zorg in range(int(le.pop())):
      n,k= list(map(int,le.pop().split()))
      #print(brute(n,k))
      if n%2:
          af.append(pow(1+pow(2,n-1,mo),k,mo))
      else:
          a=pow(2,n-1,mo)-1
          b=pow(2,n,mo)
          af.append(sum(pow(a,egaux,mo)*pow(b,max(0,k-1-egaux),mo) for egaux in range(k+1))%mo)
  print("\n".join(map(str,af)))
  