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: def ksm(a, b, c=int(1e9+7)):
      if b == 0:
          return 1
      a = a%c
      res = 1
      while b:
          if b%2 == 1:
              res = res*a%c
          a = a*a%c
          b = b // 2
      return res
  
  p = int(1e9+7)
  mi = [1]
  for i in range(211111):
      mi.append(mi[i]*2%p)
  for _ in range(int(input())):
      n, k = map(int, input().split())
      if n % 2 == 1:
          da = 0
          den0 = ksm(2, n-1)
          den1 = 1
      else:
          da = 1
          den0 = ksm(2, n-1) - 1
          den1 = 0
      res = 0
      if n % 2 == 0:
          cz0 = 1
          cz1 = ksm(2, n)
          for i in range(1, k+1):
              res = (res + ksm(cz1, k-i)*cz0)%p
              cz0 = cz0*den0 % p
      res = (res + ksm(den1+den0, k))%p
      if k == 0:
          print(1)
      else:
          print(res)
  