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: import sys
  import collections
  
  infile = sys.stdin.buffer
  def gs()  : return infile.readline().rstrip()
  def gi()  : return int(gs())
  def gss() : return gs().split()
  def gis() : return [int(x) for x in gss()]
  
  MOD = (10 ** 9) + 7
  INF = 10 ** 10
  
  p = [1 for i in range(2*(10 ** 5) + 2)]
  for i in range(1,len(p)):
      p[i] = p[i-1] * 2
      p[i] %= MOD
  
  def pow(a,n):
      return p[n]
      if n == 0: return 1
      t = pow(a,n//2)
      ans = (t * t) % MOD 
      return ans * (a if n &1 else 1) % MOD
  
  def compute(n):
      if n & 1:
          return pow(2, n-1)
      return pow(2, n-1) -1
  
  
  
  def dfs(i, k, n,s,mem):
      if i == k:
          return 1
      prev = 1
      p = pow(2,n)
      c = 1
      for i in range(k-1,-1,-1):
          cur = 0
          cur = c if n & 1 == 0 else prev
          c*= p
          c %= MOD    
          cur += (s * prev) %MOD
          cur %= MOD
          prev = cur
      return cur
  
  
  def f(n,k):
      return dfs(0 , k, n,compute(n),[])
  
  def main(infn="") :
      global infile
      infile = open(infn,"r") if infn else open(sys.argv[1],"r") if len(sys.argv) > 1 else sys.stdin
      
      ######################################################################
      cn = int(input())
      for u in range(cn):
          n,k = map(int, input().split())
          print (f(n,k))
      ######################################################################
  
  if __name__ == '__main__' :
      main()
      #sys.stdout.flush()