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
  from math import gcd
  input = sys.stdin.readline
  mod = 10**9+7
  
  t = int(input())
  for _ in range(t):
      n, k = list(map(int, input().split()))
      w0 = 1
      pow_n_low = 1
      for _ in range(n-1):
          pow_n_low *= 2
          pow_n_low %= mod
      pow_n = (pow_n_low * 2) % mod
      ans = w0
      pow_n_k_low = 1
      for i in range(1, k+1):
          if n % 2 == 0:
              ans = ans*(pow_n_low-1) + pow_n_k_low
              ans %= mod
              pow_n_k_low *= pow_n
              pow_n_k_low %= mod
          if n % 2 == 1:
              ans = ans * (pow_n_low + 1)
              ans %= mod
      print(ans)
  
  
  
  
  
  
  
  