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
  
  MAX = 2*10**5+100
  MOD = 10**9+7
  fact = [0]*MAX #fact[i]: i!
  inv = [0]*MAX #inv[i]: iの逆元
  finv = [0]*MAX #finv[i]: i!の逆元
  fact[0] = 1
  fact[1] = 1
  finv[0] = 1
  finv[1] = 1
  inv[1] = 1
      
  for i in range(2, MAX):
      fact[i] = fact[i-1]*i%MOD
      inv[i] = MOD-inv[MOD%i]*(MOD//i)%MOD
      finv[i] = finv[i-1]*inv[i]%MOD
  
  def C(n, r):
      if n<r:
          return 0
      if n<0 or r<0:
          return 0
      return fact[n]*(finv[r]*finv[n-r]%MOD)%MOD
  
  for _ in range(int(input())):
      n, k = map(int, input().split())
      dp = [[0]*2 for _ in range(k+1)]
      dp[0][0] = 1
      coef00 = 0
      
      for i in range(0, n, 2):
          coef00 += C(n, i)
          coef00 %= MOD
      
      coef01 = 0
      
      for i in range(1, n, 2):
          coef01 += C(n, i)
          coef01 %= MOD
          
      coef10 = 1 if n%2==0 else 0
      coef11 = 1 if n%2==1 else 0
      
      for i in range(k):
          #00
          dp[i+1][0] += dp[i][0]*coef00%MOD
          dp[i+1][0] %= MOD
          dp[i+1][1] += dp[i][1]*coef00%MOD
          dp[i+1][1] %= MOD
          #01
          dp[i+1][1] += dp[i][1]*coef01%MOD
          dp[i+1][1] %= MOD
          #10
          dp[i+1][1] += dp[i][0]*coef10
          dp[i+1][1] %= MOD
          dp[i+1][1] += dp[i][1]*coef10
          dp[i+1][1] %= MOD
          #11
          dp[i+1][0] += dp[i][0]*coef11
          dp[i+1][0] %= MOD
          dp[i+1][1] += dp[i][1]*coef11
          dp[i+1][1] %= MOD
      
      print(sum(dp[k])%MOD)