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 collections import Counter, defaultdict, deque
  import bisect
  from sys import stdin, stdout
  from itertools import repeat, permutations
  import math
  import heapq
  
  
  MOD = int(1e9+7)
  
  def inp(force_list=False):
      re = map(int, raw_input().split())
      if len(re) == 1 and not force_list:
          return re[0]
      return re
  
  def inpf(force_list=False):
      re = map(float, raw_input().split())
      if len(re) == 1 and not force_list:
          return re[0]
      return re
  
  def inst():
      return raw_input().strip()
  
  def gcd(x, y):
     while(y):
         x, y = y, x % y
     return x
  
  def lcm(x, y):
      return x*y/(gcd(x,y))
  
  def qmod(a, b, mod=MOD):
      res = 1
      while b:
          if b&1:
              res = (res*a)%mod
          b >>= 1
          a = (a*a)%mod
          # print b
      return res
  
  def inv(a):
      return qmod(a, MOD-2)
  
  def query(num):
      print num
      stdout.flush()
      return inp()
  
  
  def my_main():
      kase = inp()
      pans = []
      cur = []
      for _ in range(kase):
          n, k = inp()
          odd, ev = qmod(2, n-1), qmod(2, n-1)
          if n%2==1:
              odd -=1
          else:
              ev -= 1
          ans = 0
          att = qmod(2, n)
          tt = left = qmod(att, k)
          inv2n = inv(att)
          add = n%2 + ev
          pre = 1
          for i in range(k):
              left = (left * inv2n) % MOD
              ans = (ans + pre * left * odd) % MOD
              pre = (pre * add) % MOD
          print (tt-ans+MOD)%MOD
  
      # print '\n'.join(pans)
  
  my_main()
  