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: def powmod(b,e,n):
      """powmod(b,e,n) computes the eth power of b mod n.  
      (Actually, this is not needed, as pow(b,e,n) does the same thing for positive integers.
      This will be useful in future for non-integers or inverses."""
      accum = 1; i = 0; bpow2 = b
      while ((e>>i)>0):
          if((e>>i) & 1):
              accum = (accum*bpow2) % n
          bpow2 = (bpow2*bpow2) % n
          i+=1
      return accum
  
  def memoize(function):
      cache = {}
      def decorated_function(*args):
          try:
              return cache[args]
          except KeyError:
              val = function(*args)
              cache[args] = val
              return val
      return decorated_function
  
  @memoize
  def inverseMod(n, mod):
      if 1 == n: return 1
  
      return ((mod - mod//n) * inverseMod(mod % n, mod)) % mod
  
  M = 10**9 + 7
  
  F = [1, 1]
  for n in range(2, 2*10**5 + 1):
      F.append((F[-1] * n) % M)
  
  RF = [1, 1]
  for n in range(2, 2*10**5 + 1):
      RF.append((RF[-1] * inverseMod(n, M)) % M)
  
  def C(n, k):
      return (F[n] * ((RF[n - k] * RF[k]) % M)) % M
  
  def count(n, k):
      if k == 0:
          return 1
      if n == 1:
          return powmod(2, k, M)
  
  #    r0 = 0
  #    for i in range(0, n, 2):
  #        r0 += C(n, i)
  #        r0 %= M
      r0 = powmod(2, n - 1, M) + (1 if n % 2 == 1 else 0) - 1
  
      r = 1 + r0
  
      b = powmod(2, n, M)
      bn = b
      for e in range(2, k + 1):
          if n % 2 == 0:
  #            r = ((r0 * r) % M) + powmod(2, (e - 1)*n, M)
              r = ((r0 * r) % M) + bn
              bn = (bn * b) % M
          else:
              r = r + (r0 * r) % M
          r %= M
  
      return r % M
  
  T = int(input())
  for _ in range(T):
      n, k = map(int, input().split(' '))
      print(count(n, k))
  