Problem: Pink Floyd are pulling a prank on Roger Waters. They know he doesn't like [walls](https://www.youtube.com/watch?v=YR5ApYxkU-U), he wants to be able to walk freely, so they are blocking him from exiting his room which can be seen as a grid.

Roger Waters has a square grid of size n× n and he wants to traverse his grid from the upper left (1,1) corner to the lower right corner (n,n). Waters can move from a square to any other square adjacent by a side, as long as he is still in the grid. Also except for the cells (1,1) and (n,n) every cell has a value 0 or 1 in it.

Before starting his traversal he will pick either a 0 or a 1 and will be able to only go to cells values in which are equal to the digit he chose. The starting and finishing cells (1,1) and (n,n) are exempt from this rule, he may go through them regardless of picked digit. Because of this the cell (1,1) takes value the letter 'S' and the cell (n,n) takes value the letter 'F'.

For example, in the first example test case, he can go from (1, 1) to (n, n) by using the zeroes on this path: (1, 1), (2, 1), (2, 2), (2, 3), (3, 3), (3, 4), (4, 4)

The rest of the band (Pink Floyd) wants Waters to not be able to do his traversal, so while he is not looking they will invert at most two cells in the grid (from 0 to 1 or vice versa). They are afraid they will not be quick enough and asked for your help in choosing the cells.  Note that you cannot invert cells (1, 1) and (n, n).

We can show that there always exists a solution for the given constraints.

Also note that Waters will pick his digit of the traversal after the band has changed his grid, so he must not be able to reach (n,n) no matter what digit he picks.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50). Description of the test cases follows.

The first line of each test case contains one integers n (3 ≤ n ≤ 200).

The following n lines of each test case contain the binary grid, square (1, 1) being colored in 'S' and square (n, n) being colored in 'F'.

The sum of values of n doesn't exceed 200.

Output

For each test case output on the first line an integer c (0 ≤ c ≤ 2) — the number of inverted cells.

In i-th of the following c lines, print the coordinates of the i-th cell you inverted. You may not invert the same cell twice.  Note that you cannot invert cells (1, 1) and (n, n).

Example

Input


3
4
S010
0001
1000
111F
3
S10
101
01F
5
S0101
00000
01111
11111
0001F


Output


1
3 4
2
1 2
2 1
0

Note

For the first test case, after inverting the cell, we get the following grid:
    
    
      
    S010  
    0001  
    1001  
    111F  
    

Solution (in PYTHON3):

Solution:
# cook your dish here
# cook your dish here
#from functools import reduce

#mod=int(1e9+7)
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**26)
"""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 collections import deque, defaultdict, Counter, OrderedDict
#from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, gcd
#from heapq import heappush, heappop, heapify, nlargest, nsmallest


# sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect            #c++ upperbound
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
from collections import Counter
from math import sqrt
import collections
import math
import heapq
import re
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
  
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l

def most_frequent(list):
    return max(set(list), key = list.count) 
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x

def ncr(n,r,p):                                              #To use this, Uncomment 19-25 
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t

def Convert(string): 
    li = list(string.split("")) 
    return li 

def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
prime=[]
q=[]
def dfs(n,d,v,c):
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=c
    for i in x:
        if i not in v:
            f=dfs(i,d,v,c+1)
            j=max(j,f)
            # print(f)
    return j



#Implement heapq
#grades = [110, 25, 38, 49, 20, 95, 33, 87, 80, 90] 
#print(heapq.nlargest(3, grades)) #top 3 largest
#print(heapq.nsmallest(4, grades))
#Always make a variable of predefined function for  ex- fn=len
#n,k=map(int,input().split())
"""*******************************************************"""
def main():
    for _ in range(int(input())):
        n=inin()
        a=[]
        z1=0
        o1=0
        z2=0
        o2=0
        o=[]
        z=[]
        x=[]
        for i in range(n):
            a.append(sin())
        if a[0][1] == '0':
            z1+=1
            x.append(1)
        else:
            o1+=1
        if a[1][0]=='0':
            z1+=1
            x.append(2)
        else:
            o1+=1
        if a[n-2][n-1] == '0':
            z2+=1
            z.append(1)
        else:
            o2+=1
            o.append(1)
        if a[n-1][n-2]=='0':
            z2+=1
            z.append(2)
        else:
            o2+=1      
            o.append(2)      
        if (z1==0 and o2==0) or (z2==0 and o1==0):
            print('0')
            continue 
        if z1==2:
            if z2==2:
                print('2')
                print(n,n-1)
                print(n-1,n)
            elif z2==1:
                print('1')
                if z[-1]==2:
                    print(n,n-1)
                else:
                    print(n-1,n)
            continue
        if o1==2:
            if o2==2:
                print('2')
                print(n,n-1)
                print(n-1,n)
            elif z2==1:
                print('1')
                if o[-1]==2:
                    print(n,n-1)
                else:
                    print(n-1,n)
            continue
        if o1==1:
            if z2==1:
                print(2)
                if x[-1]==1:
                    print(1,2)
                else:
                    print(2,1)
                if o[-1]==1:
                    print(n-1,n)
                else:
                    print(n,n-1)
            else:
                if z2==0:
                    print(1)
                    if x[-1]==1:
                        print(2,1)
                    else:
                        print(1,2)
                else:
                    print(1)
                    if x[-1]==1:
                        print(1,2)
                    else:
                        print(2,1)














    
"""*******************************************************"""
######## Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
if __name__== "__main__":
    main()
#threading.Thread(target=main).start()