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 collections import Counter
  import string
  import math
  import bisect
  #import random
  import sys
  # sys.setrecursionlimit(10**6) 
  from fractions import Fraction
  def modDivide(a,b,m):
      inv = pow(b, m - 2, m)
      return (inv*a) % m
  def array_int():
      return [int(i) for i in sys.stdin.readline().split()]
  def vary(arrber_of_variables):
      if arrber_of_variables==1:
          return int(sys.stdin.readline())
      if arrber_of_variables>=2:
          return map(int,sys.stdin.readline().split()) 
  def makedict(var):
      return dict(Counter(var))
  testcases=vary(1)
  for _ in range(testcases):
      n,k=vary(2)
      mod=10**9+7
      if k==0:
          print(1)
          continue 
      if(n%2==1):
          print(pow(1+pow(2,n-1,mod),k,mod))
      else:
          t = pow(pow(2,n-1,mod)-1,k,mod)
          a = pow(2,n*k,mod)
          b = (pow(2,n-1,mod)+1)%mod
          a = modDivide((a-t+mod)%mod,b,mod)
          t=(t+a)%mod
          print(t)
  