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: MOD = 1000000007
  
  
  def qmod(a, b, mod=MOD):
      res = 1
      while b:
          if b & 1:
              res = (res*a) % mod
          b = b >> 1
          a = (a*a) % mod
          # print b
      return res
  
  
  def inv(a):
      return qmod(a, MOD-2)
  
  
  t = int(raw_input())
  
  for zz in range(t):
      n, k = raw_input().split(' ')
      n, k = int(n), int(k)
      odd, ev = qmod(2, n - 1), qmod(2, n - 1)
      if n % 2 == 1:
          odd -= 1
      else:
          ev -= 1
      ans = 0
      att = qmod(2, n)
      tt = left = qmod(att, k)
      inv2n = inv(att)
      add = n % 2 + ev
      pre = 1
      for i in range(k):
          left = (left * inv2n) % MOD
          ans = (ans + pre * left * odd) % MOD
          pre = (pre * add) % MOD
      print (tt - ans + MOD) % MOD
  