1557_C. Moamen and XOR_183
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].
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat, permutations
import math
import heapq


MOD = int(1e9+7)

def inp(force_list=False):
    re = map(int, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inpf(force_list=False):
    re = map(float, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inst():
    return raw_input().strip()

def gcd(x, y):
   while(y):
       x, y = y, x % y
   return x

def lcm(x, y):
    return x*y/(gcd(x,y))

def qmod(a, b, mod=MOD):
    res = 1
    while b:
        if b&1:
            res = (res*a)%mod
        b >>= 1
        a = (a*a)%mod
        # print b
    return res

def inv(a):
    return qmod(a, MOD-2)

def query(num):
    print num
    stdout.flush()
    return inp()


def my_main():
    kase = inp()
    pans = []
    cur = []
    for _ in range(kase):
        n, k = inp()
        odd, ev = qmod(2, n-1), qmod(2, n-1)
        if n%2==1:
            odd -=1
        else:
            ev -= 1
        ans = 0
        att = qmod(2, n)
        tt = left = qmod(att, k)
        inv2n = inv(att)
        add = n%2 + ev
        pre = 1
        for i in range(k):
            left = (left * inv2n) % MOD
            ans = (ans + pre * left * odd) % MOD
            pre = (pre * add) % MOD
        print (tt-ans+MOD)%MOD

    # print '\n'.join(pans)

my_main()
