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: real_ans=0
  def abc(x,n,k):
      global real_ans
      if len(x)==n:
          a=x[0]
          b=x[0]
          
          for j in x[1:]:
              a&=j
              b^=j
          
          if a>=b:
              real_ans+=1
          return
      for j in range(2**k):
          abc(x+[j],n,k)
      
  mod =10**9+7
  e=2*10**5+1
  fact=[1]*e
  for j in range(1,e):
      fact[j] =(j*fact[j-1])%mod
  def ncr(a,b):
      if a<=0 or a<b:
          return 0
      x = fact[a]
      y = pow(fact[b]*fact[a-b],mod-2,mod)
      return (x*y)%mod
   
  t=int(input())
  for _ in range(t):
      n,k=map(int,input().split())
      
      if k==0:
          print(1)
      else:
          b=pow(2,n-1,mod)
          d=1
          e=pow(2,n,mod)
          ans=1
          for j in range(k):
              y=ans
              if n%2:
                  ans=(ans+y)%mod
              else:
                  ans=(ans+d)%mod
              #x=2
              #while x<n:
                  #ans=(ans+(ncr(n,x)*y))%mod
                  #x+=2
              ans=(ans+b*y-y-y*(1 if n%2==0 else 0))%mod
              d=(d*e)%mod
              
          
          print(ans%mod)
      
  