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: inf = float('inf')
  mod = 10 ** 9 + 7
  def get_list(): return [int(i) for i in input().split()]
  def yn(a): print("YES" if a else "NO")
  ceil = lambda a, b: (a + b - 1) // b
  t=int(input())
  for i in range(t):
      n,k=[int(i) for i in input().split()]
      dp=[1 for i in range(k+1)]
      muleven=pow(2,n-1,mod)
      twon=pow(2,n,mod)
      mulodd=pow(2,n-1,mod)+1
      if n%2==0:
          for i in range(1,k+1):
              dp[i]=(dp[i-1]*(muleven-1))%mod
              dp[i]=(pow(twon,i-1,mod)+dp[i])%mod
          print(dp[-1])
      else:
          for i in range(1,k+1):
              dp[i]=(dp[i-1]*mulodd)%mod
          print(dp[-1])