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 os.path
  from collections import *
  import math
  import bisect
  import heapq as hq
  from fractions import Fraction
  from random import randint
  
  if (os.path.exists('input.txt')):
      sys.stdin = open("input.txt", "r")
      sys.stdout = open("output.txt", "w")
  
  mod = 10 ** 9 + 7
  
  def modExp(x, y):
      if y == 0:
          return 1
      if y % 2 == 0:
          return modExp((x * x) % mod, y // 2)
      return (x * modExp((x * x) % mod, (y - 1) // 2)) % mod
  
  ##########################################################
  
  
  t = int(input())
  
  while t:
      t -= 1
      n, k = [int(x) for x in input().split()]
      if k == 0:
          print(1)
          continue
      res = 0
      z = modExp(2,n - 1)
      y = z * 2
      p = 1
      if n % 2 == 0:
          for i in range(k):
              x = 1
              x = (x * modExp(y, (k - i - 1))) % mod
              x = (x * p) % mod
              res = (res + x) % mod
              p = (p * (z - 1)) % mod
          res = (res + p) % mod
      else:
          res = modExp(z + 1,k)
      print(res)
  
  ##########################################################
  