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: #!/usr/bin/env python3
  # from typing import *
  
  import sys
  import io
  import math
  import collections
  import decimal
  import itertools
  import bisect
  import heapq
  
  
  def input():
      return sys.stdin.readline()[:-1]
  
  
  # sys.setrecursionlimit(1000000)
  
  # _INPUT = """3
  # 4 2
  # 2 1
  # 4 0
  # """
  # sys.stdin = io.StringIO(_INPUT)
  
  INF = 10**10
  
  MOD = 1000000007
  
  def solve(N, K):
      if K == 0:
          return 1
  
      D = 0
      for i in range(0, N, 2):
          D = (D + comb(N, i)) % MOD
  
      if N % 2 == 1:
          ans = pow(D+1, K, MOD)
          return ans
  
      else:
          D_inv = pow(D, -1+(MOD-1), MOD) #pow(D, -1, MOD)
          n_any = pow(2, N, MOD)
          v = pow(D, K, MOD)
          ans = v
          v = (v * D_inv) % MOD
          ans = (ans + v) % MOD
          for _ in range(K-1):
              v = (v * D_inv * n_any) % MOD
              ans = (ans + v) % MOD
          return ans
  
  fact = [1, 1]  # fact[n] = (n! mod p)
  factinv = [1, 1]  # factinv[n] = ((n!)^(-1) mod p)
  inv = [0, 1]  # factinv 計算用
  
  def comb(n, r):
      r = min(r, n - r)
      return (fact[n] * factinv[r] * factinv[n-r]) % MOD
  
  N_MAX = 2*(10**5)
  for i in range(2, N_MAX + 1):
      fact.append((fact[-1] * i) % MOD)
      inv.append((-inv[MOD % i] * (MOD // i)) % MOD)
      factinv.append((factinv[-1] * inv[-1]) % MOD)
  
  T0 = int(input())
  for _ in range(T0):
      N, K = map(int, input().split())
      print(solve(N, K))