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 comb(n,r,mod):
      if r<0 or r>n:return 0
      return g1[n]*g2[r]*g2[n-r]%mod
  mod=10**9+7
  N=2*10**5+1
  g1=[1,1]
  g2=[1,1]
  inv=[0,1]
  for i in range(2,N+1):
      g1.append((g1[-1]*i)%mod)
      inv.append((-inv[mod%i]*(mod//i))%mod)
      g2.append((g2[-1]*inv[-1])%mod)
  t=int(input())
  for _ in range(t):
      n,k=map(int,input().split())
      dp=[[0]*2 for i in range(k+1)]
      dp[0][0]=1
      p2=pow(2,n,mod)
      s=0
      for j in range(0,n,2):
          s+=comb(n,j,mod)
          s%=mod
      if n%2:s+=1
      for i in range(1,k+1):
          if n%2==0:
              dp[i][1]=(dp[i-1][1]*p2+dp[i-1][0])%mod
          else:
              dp[i][1]+=dp[i-1][1]*p2%mod
          dp[i][0]=dp[i-1][0]*s%mod
      print(sum(dp[k])%mod)