human
Ashish and Vivek play a game on a matrix consisting of n rows and m columns, where they take turns claiming cells. Unclaimed cells are represented by 0, while claimed cells are represented by 1. The initial state of the matrix is given. There can be some claimed cells in the initial state.

In each turn, a player must claim a cell. A cell may be claimed if it is unclaimed and does not share a row or column with any other already claimed cells. When a player is unable to make a move, he loses and the game ends.

If Ashish and Vivek take turns to move and Ashish goes first, determine the winner of the game if both of them are playing optimally.

Optimal play between two players means that both players choose the best possible strategy to achieve the best possible outcome for themselves.

Input

The first line consists of a single integer t (1 ≤ t ≤ 50) — the number of test cases. The description of the test cases follows.

The first line of each test case consists of two space-separated integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the matrix.

The following n lines consist of m integers each, the j-th integer on the i-th line denoting a_{i,j} (a_{i,j} ∈ \{0, 1\}).

Output

For each test case if Ashish wins the game print "Ashish" otherwise print "Vivek" (without quotes).

Example

Input

4
2 2
0 0
0 0
2 2
0 0
0 1
2 3
1 0 1
1 1 0
3 3
1 0 0
0 0 0
1 0 0

Output

Vivek
Ashish
Vivek
Ashish

Note

For the first case: One possible scenario could be: Ashish claims cell (1, 1), Vivek then claims cell (2, 2). Ashish can neither claim cell (1, 2), nor cell (2, 1) as cells (1, 1) and (2, 2) are already claimed. Thus Ashish loses. It can be shown that no matter what Ashish plays in this case, Vivek will win. 

For the second case: Ashish claims cell (1, 1), the only cell that can be claimed in the first move. After that Vivek has no moves left.

For the third case: Ashish cannot make a move, so Vivek wins.

For the fourth case: If Ashish claims cell (2, 3), Vivek will have no moves left.
Using python can you solve the prior task?
 ... Write me a Python program to determine the winner of the game if both of them are playing optimally?
assistant
#Here is Python program to determine the winner of the game if both of them are playing optimally.

#from __future__ import print_function
mod=int(1e9+7)
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**25)
#import sys
#sys.setrecursionlimit(10**5)
#fact=[1]
#for i in range(1,100001):
#    fact.append((fact[-1]*i)%mod)
#ifact=[0]*100001
#ifact[100000]=pow(fact[100000],mod-2,mod)
#for i in range(100000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
#from random import randint as rn
import bisect
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
import collections
import math
import heapq
from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):                        #for using this uncomment the lines calculating fact and ifact
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def val(x):
    if(x<=0):
        return 0
    return (x*(x+1))/2
range=xrange
"""*******************************************************"""
def main():
    for _ in range(int(sin())):
        n,m=ain()
        a=[]
        for i in range(n):
            a.append(ain())
        r=0
        c=0
        for i in range(n):
            f=0
            for j in range(m):
               ...