human
PYTHON solution for "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].":


 Can you provide a Python code to print the result modulo 1 000 000 007 (10^9 + 7)?
assistant
#Here is Python code to print the result modulo 1 000 000 007 (10^9 + 7).
 from __future__ import division,print_function
from heapq import*
import sys
le = sys.__stdin__.read().split("\n")[::-1]
af=[]
mo=10**9+7
def brute(n,k):
    l=[]
    def gen(n,k,ind=0):
        if ind!=n:
            s=0
            for i in range(2**k):
                l.append(i)
                s+=gen(n,k,ind+1)
                l.pop()
            return s
        else:
            an=l[0]
            xo=0
            for i in l:
                an &= i
                xo ^= i
            return an>=xo
    return gen(n,k)
for zorg in range(int(le.pop())):
    n,k= list(map(int,le.pop().split()))
    #print(brute(n,k))
    if n%2:
        af.append(pow(1+pow(2,n-1,mo),k,mo))
    else:
        a=pow(2,n-1,mo)-1
        b=pow(2,n,mo)
        af.append(sum(pow(a,egaux,mo)*pow(b,max(0,k-1-egaux),mo) for egaux in range(k+1))%mo)
print("\n".join(map(str,af)))