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 math import *
  import sys
  input = sys.stdin.readline
  print = sys.stdout.write
  t = int(input())
  mod = 1000000007
  
  def power(x, y):
      x %= mod
      res = 1
      while (y > 0):
          if (y & 1):
              res *= x
              res %= mod
          y >>= 1
          x *= x
          x %= mod
      return res
  
  for q in range(t):
      dp = []
      z = 1
      dp.append(1)
      s = input().split()
      n = int(s[0])
      k = int(s[1])
      for i in range(1, k + 1):
          dp.append(0)
          if n & 1:
              dp[i] = ((power(2, n - 1) * dp[i - 1]) % mod + dp[i - 1]) % mod
          else:
              dp[i] = (z + ((power(2, n - 1) - 1) * dp[i - 1])) % mod
          z *= power(2, n)
          z %= mod
      print(str(int(dp[k])) + '\n')