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 heapq;import sys;import math;from collections import deque;from bisect import bisect_left,bisect_right;I=sys.stdin.readline;II=lambda :int(I());IN=lambda x:map(int,x.split());FN=lambda x:map(float,x.split());L=lambda x:list(IN(x));M=1000000007;P=print;T=True;F=False
  for _ in range(II()):
  	n,k=IN(I())
  	tot=pow(2,n-1,M);g=0;eq=0;st=0
  	if(n&1==1):
  		eq=(tot+1)%M
  	else:
  		g=1;st=(tot*2)%M
  		eq=(tot-1)%M
  	res=0;t=1
  	powers=[1]
  	if(g==1):
  		i=1
  		while(i<k):
  			powers.append((powers[-1]*st)%M)
  			i+=1
  	i=k-1
  	while(i>=0):
  		if(g==1):
  			res=(res+(powers[i]*t)%M)%M
  		t=(t*eq)%M
  		i-=1
  	res=(res+t)%M
  	print(res)
  