1557_C. Moamen and XOR_185
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].
import sys
from sys import stdout

input = lambda: sys.stdin.readline().strip()
P = lambda: list(map(int, input().split()))
from math import factorial as f, gcd, ceil
from collections import deque, defaultdict as dd, Counter as C
from heapq import heapify, heappop, heappush, heappushpop, heapreplace, merge
from random import randint, choice, sample
from operator import itemgetter as ig, attrgetter as ag, methodcaller as mc
import time
mod = 10**9+7
a = ord('a')

# Fenwick Tree
class FT:
    def __init__(self, N):
        self.ft = [0] * (N+1)
        self.N = N

    def ls(self, x):
        return x & -x

    def fenwick_update(self, p, v):
        while p <= self.N:
            self.ft[p] += v
            p += self.ls(p)

    def fenwick_query(self, p):
        s = 0
        while p:
            s += self.ft[p]
            p -= self.ls(p)
        # print(p, s)
        return s

start = time.time()
def fast_exp(x, exp):
    ans = 1
    base = x
    while exp:
        if exp & 1:
            ans *= base
        base *= base
        base %= mod
        ans %= mod
        exp >>= 1
    return ans

def countBits(n):
    count = 0
    while n:
        count += n & 1
        n >>= 1
    return count

def submasks(n):
    #this is cool
    #https://cp-algorithms.com/algebra/all-submasks.html
    org = n
    while n:
        yield n
        n = (n-1) & org


def solve():
    n, k = P()
    if k == 0:
        print(1)
    elif n % 2:
        # print((1 << (k*(n-1))) % mod)
        print(fast_exp(fast_exp(2, n-1) + 1, k))
    else:
        z = fast_exp(2, n-1) - 1
        z1 = fast_exp(2, n)
        ans = 0
        # left = [1] * k
        right = [1] * k
        # for i in range(1, k):
        #     left[i] = (left[i-1] * z) % mod
        for i in range(k-2, -1, -1):
            right[i] = (right[i+1] * z1) % mod
        left = 1
        for i in range(k):
            # print(left, right)
            # ans = (ans + (1<<((n-1)*left + n*right))) % mod
            ans = (ans + left*right[i]) % mod
            left = (left * z) % mod
            # ans = (ans + (1<<(n*right))) % mod
        # print((ans + (1 << ((n-1)*k))) % mod)
        # print(ans)
        # print((ans + fast_exp(2, (n-1)*k)) % mod)
        # print(ans)

        print((ans + fast_exp(z, k) ) % mod)
    # print('--')


tc = int(input())
for t in range(1, tc+1):
    solve()

# solve()
# print(time.time()-start)


