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
  input = sys.stdin.readline
  def mp():return map(int,input().split())
  def lmp():return list(map(int,input().split()))
  def mps(A):return [tuple(map(int, input().split())) for _ in range(A)]
  import math
  import bisect
  from copy import deepcopy as dc
  from itertools import accumulate
  from collections import Counter, defaultdict, deque
  def ceil(U,V):return (U+V-1)//V
  def modf1(N,MOD):return (N-1)%MOD+1
  inf = int(1e20)
  mod = int(1e9+7)
  
  t = int(input())
  for _ in range(t):
      n,k = mp()
      if k == 0:
          print(1)
          continue
      if n % 2 == 1:
          ans = 1
          for i in range(k):
              ans *= (pow(2,n-1,mod)+1)
              ans %= mod
      else:
          ans = 0
          u = pow(2,n,mod)
          v = pow(2,n-1,mod)-1
          for i in range(k):
              ans += pow(u,k-i-1,mod) * pow(v,i,mod)
              ans %= mod
          ans += pow(v,k,mod)
      print(ans%mod)
  
  
  
  
  