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 collections
  # import random
  # import math
  #import itertools
  #import math
  #mport math
  from collections import defaultdict
  # import itertools
  # from sys import stdin, stdout
  #import math
  import sys
  # import operator
  # from decimal import Decimal
  
  # sys.setrecursionlimit(10**6)
  
  p2D = lambda x: print(*x, sep="\n")
  def II(): return int(sys.stdin.buffer.readline())
  def MI(): return map(int, sys.stdin.buffer.readline().split())
  def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def BI(): return sys.stdin.buffer.readline().rstrip()
  def SI(): return sys.stdin.buffer.readline().rstrip().decode()
  def li(): return [int(i) for i in input().split()]
  def lli(rows): return [li() for _ in range(rows)]
  def si(): return input()
  def ii(): return int(input())
  def ins(): return input().split()
  
  
  def solve():
      n, k = LI()
      mod = 10**9+7
      if k == 0:
          return 1
      else:
          sbd = pow(2,n-1, mod)
          ap = (sbd*2)%mod
          if n%2:
              draw = pow(sbd+1,k,mod)
              win=0
          else:
              draw = pow(sbd - 1, k, mod)
              # win = (((pow(ap,k,mod)- pow(sbd-1,k,mod))%mod) // ((sbd+1)%mod))
              #print(sbd, ap)
              win = 0
              apf = 1
              sbdf = [1]
              sbd-=1
              for l in range(k-1):
                  sbdf.append((sbdf[-1]*sbd)%mod)
              for j in range(k):
                  win= (win+ (apf * sbdf[k-j-1])%mod)%mod
                  apf = (ap * apf) % mod
  
          return (win+draw)%mod
  
  
  
  def main():
      for _ in range(II()):
          sys.stdout.write(str(solve()) + "\n")
  
  
  
      #     z += str(ans) + '\n'
      # print(len(ans), ' '.join(map(str, ans)), sep='\n')
      # stdout.write(z)
  
  
  # for interactive problems
  # print("? {} {}".format(l,m), flush=True)
  # or print this after each print statement
  # sys.stdout.flush()
  
  
  if __name__ == "__main__":
      main()