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: p = 10**9+7
  def power(x, y, p):
      b = bin(y)[2:]
      start = x % p
      answer =1 
      for i in range(len(b)):
          if b[len(b)-1-i]=='1':
              answer = (answer*start) % p
          start = (start*start) % p
      return answer
  
  def invert(x, p):
      return power(x, p-2, p)
  
  
  def process(n, k):
      a1 = power(2, n*k, p)
      if n % 2==0:
          a2 = (power(2, n-1, p) % p)
          a3 = (a2-1) % p
          R = (invert(2, p)-invert(power(2, n, p), p)) % p
          num1 = (power(R, k+1, p)-R) % p
          den1 = (R-1) % p
          num = (a2*a1*num1) % p
          den = (a3*den1) % p
          entry = (num*invert(den, p)) % p
          return (a1-entry) % p
      else:
          a2 = (power(2, n-1, p)-1) % p
          a3 = (a2+2) % p
          R = (invert(2, p)+invert(power(2, n, p), p)) % p
          num1 = (power(R, k+1, p)-R) % p
          den1 = (R-1) % p
          num = (a2*a1*num1) % p
          den = (a3*den1) % p
          entry = (num*invert(den, p)) % p
          return (a1-entry) % p
      
  t = int(input())
  for i in range(t):
      n,k = [int(x) for x in input().split()]
      print(process(n, k))