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: # ------------------- fast io --------------------
  import os
  import sys
  from io import BytesIO, IOBase
  
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
      
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
      
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
      
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
      
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  # ------------------- fast io --------------------
  from math import ceil
  
  
  def prod(a, mod=10 ** 9 + 7):
      ans = 1
      for each in a:
          ans = (ans * each) % mod
      return ans
  
  
  def gcd(x, y):
      while y:
          x, y = y, x % y
      return x
  
  
  def lcm(a, b): return a * b // gcd(a, b)
  
  
  def binary(x, length=16):
      y = bin(x)[2:]
      return y if len(y) >= length else "0" * (length - len(y)) + y
  
  
  for _ in range(int(input()) if True else 1):
      #n = int(input())
      n, k = map(int, input().split())
      # a, b = map(int, input().split())
      # c, d = map(int, input().split())
      # a = list(map(int, input().split()))
      # b = list(map(int, input().split()))
      # s = input()
      mod = 10**9 + 7
      if n % 2:
          print(pow(pow(4, n//2, mod)+1, k, mod))
      else:
          p = pow(2, n - 1, mod)
          q = p * (p + 1)
          x = 6 * pow(4, (n//2 - 1), mod) - 1
          y = p*x - q
          a = [1, p, q]
          while len(a) <= k + 2:
              a += [x*a[-1] - y*a[-2]]
              a[-1] = a[-1] % mod
          print(a[k] % mod)