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: from sys import stdin, stdout
  m = 1000000007
  
  twotothe = [1]
  facts = [1]
  for i in range(1, 200001):
      facts.append((facts[-1]*i)%m)
      twotothe.append((twotothe[-1]*2)%m)
  
  invnums = [0]*200001
  invfacts = [0]*200001
  
  invnums[0] = 1
  invnums[1] = 1
  for i in range(2, 200001):
      invnums[i] = (invnums[m%i] * (m - m//i))%m
  
  invfacts[0] = 1
  invfacts[1] = 1
  for i in range(2, 200001):
      invfacts[i] = (invnums[i] * invfacts[i-1]) % m
  
  def comb(n, r):
      return (facts[n]*invfacts[n-r]*invfacts[r])%m
  
  for _ in range(int(stdin.readline())):
      n, k = map(int, stdin.readline().split())
      if k == 0:
          stdout.write("1\n")
          continue
      sm = sum(comb(n, r) for r in range(0, n, 2))%m
      if n&1:
          stdout.write(str(pow(sm+1, k, m))+'\n')
          continue
      ans = 1
      last = 1
      for i in range(k):
          ans = (last+sm*ans)%m
          last = (last*twotothe[n])%m
      stdout.write(str(ans)+'\n')
  