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 io, os
  import math
  from heapq import *
  gcd = math.gcd
  sqrt = math.sqrt
  def ceil(a,b):
      a=-a
      k=a//b
      k=-k
      return k
  # arr=list(map(int, input().split()))
  input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  def strinp(testcases):
      k = 5
      if (testcases == -1 or testcases == 1):
          k = 1
      f = str(input())
      f = f[2:len(f) - k]
      return f
  p=1000000007
  def fun(n,k):
      f=pow(2,n-1,p)
      for i in range(2,k+1):
          f=(pow(pow(2,n,p),i-1,p)+(((pow(2,n-1,p)-1)%p)*f)%p)%p
      return f
  def main():
      t=int(input())
      for _ in range(t):
          arr = list(map(int, input().split()))
          n=arr[0]
          k=arr[1]
          if(k==0):
              print(1)
              continue
          if(n%2==1):
              a=pow(2,n-1,p)
              a+=1
              a=a%p
              a=pow(a,k,p)
              print(a)
          else:
              print(fun(n,k))
  main()