Problem: This is the easy version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.

You are given a binary table of size n × m. This table consists of symbols 0 and 1.

You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).

Your task is to make all symbols in the table equal to 0. You are allowed to make at most 3nm operations. You don't need to minimize the number of operations.

It can be proved that it is always possible.

Input

The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.

The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).

Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.

It is guaranteed that the sum of nm for all test cases does not exceed 20000.

Output

For each test case print the integer k (0 ≤ k ≤ 3nm) — the number of operations.

In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong into some 2 × 2 square.

Example

Input


5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101


Output


1
1 1 2 1 2 2
2 
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2 
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2 
1 4 1 5 2 5 
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2

Note

In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.

In the second test case:

  * operation with cells (2, 1), (3, 1), (3, 2). After it the table will be: 
    
          
    011  
    001  
    000  
    

  * operation with cells (1, 2), (1, 3), (2, 3). After it the table will be: 
    
          
    000  
    000  
    000  
    




In the fifth test case:

  * operation with cells (1, 3), (2, 2), (2, 3). After it the table will be: 
    
          
    010  
    110  
    

  * operation with cells (1, 2), (2, 1), (2, 2). After it the table will be: 
    
          
    000  
    000  
    

Solution (in PYTHON3):

Solution:
import sys, os
import heapq, functools, collections
import math, random
from collections import Counter, defaultdict

# available on Google, not available on Codeforces
# import numpy as np
# import scipy


def solve_(g,n,m):
    og = [[x for x in row] for row in grid]
    # your solution here

    x,y = 0,0
    
    res = []

    # leave the last two rows out
    # crawl based on top row

    for y in range(n-2):  # special consideration for last two rows
        for x in range(m-1):  # special consideration for last column
            if g[y][x] == 0:
                continue

            if g[y][x] == 1 and g[y][x+1] == 1:
                if g[y+1][x] == 1:
                    r = [(y,x), (y,x+1), (y+1,x)]
                else:
                    r = [(y,x), (y,x+1), (y+1,x+1)]
            if g[y][x] == 1 and g[y][x+1] == 0:
                r = [(y,x), (y+1,x), (y+1,x+1)]

            #
            assert len(r) == 3
            assert len(set(r)) == 3
            res.append(r)
            for yy,xx in r:
                g[yy][xx] = 1-g[yy][xx]
            #

        assert x+1 == m-1
        if g[y][x] == 0 and g[y][x+1] == 0:
            continue
        elif g[y][x] == 1 and g[y][x+1] == 0:
            r = [(y,x),   (y+1,x), (y+1,x+1)]
        elif g[y][x] == 0 and g[y][x+1] == 1:
            r = [(y,x+1), (y+1,x), (y+1,x+1)]
        elif g[y][x] == 1 and g[y][x+1] == 1:
            r = [(y,x), (y,x+1), (y+1,x)]
        
        #
        assert len(r) == 3
        assert len(set(r)) == 3
        res.append(r)
        for yy,xx in r:
            g[yy][xx] = 1-g[yy][xx]
        #

    # console(y, n-3)
    # assert y == n-3
    for x in range(m-2):
        if g[-2][x] == 0 and g[-1][x] == 0:
            continue
        if g[-2][x] == 1 and g[-1][x] == 0:
            r = [(-2,x), (-2,x+1), (-1,x+1)]
        if g[-2][x] == 0 and g[-1][x] == 1:
            r = [(-1,x), (-2,x+1), (-1,x+1)]
        if g[-2][x] == 1 and g[-1][x] == 1:
            if g[-2][x+1] == 1:
                r = [(-2,x), (-1,x), (-2,x+1)]
            else:
                r = [(-2,x), (-1,x), (-1,x+1)]
        #
        assert len(r) == 3
        assert len(set(r)) == 3
        res.append(r)
        for yy,xx in r:
            g[yy][xx] = 1-g[yy][xx]
        #

    l4 = [(-2,-2), (-2,-1), (-1,-2), (-1,-1)]

    if g[-2][-2] + g[-2][-1] + g[-1][-2] + g[-1][-1] == 4:
        console("resolve 4")
        r = l4[:3]
        #
        assert len(r) == 3
        assert len(set(r)) == 3
        res.append(r)
        for yy,xx in r:
            g[yy][xx] = 1-g[yy][xx]
        # console(grid)
        #

    if g[-2][-2] + g[-2][-1] + g[-1][-2] + g[-1][-1] == 1:
        console("resolve 1")
        for i,(y,x) in enumerate(l4 + l4):
            if g[y][x] == 1:
                r = (l4+l4)[i:i+3]
                break
        #
        assert len(r) == 3
        assert len(set(r)) == 3
        res.append(r)
        for yy,xx in r:
            g[yy][xx] = 1-g[yy][xx]
        # console(grid)
        #

    if g[-2][-2] + g[-2][-1] + g[-1][-2] + g[-1][-1] == 2:
        console("resolve 2")
        sett = set()
        for i,(y,x) in enumerate(l4 + l4 + l4):
            if g[y][x] == 0:
                sett.add((y,x))
            if len(sett) == 2:
                sett.add((y,x))
            if len(sett) == 3:
                break
        r = list(sett)

        #
        assert len(r) == 3
        assert len(set(r)) == 3
        res.append(r)
        for yy,xx in r:
            g[yy][xx] = 1-g[yy][xx]
        #
        # console(grid)


    if g[-2][-2] + g[-2][-1] + g[-1][-2] + g[-1][-1] == 3:
        console("resolve 3")
        r = []
        for i,(y,x) in enumerate(l4):
            if g[y][x] == 1:
                r.append((y,x))

        #
        assert len(r) == 3
        assert len(set(r)) == 3
        res.append(r)
        for yy,xx in r:
            g[yy][xx] = 1-g[yy][xx]
        #

    # console(res)
    res2 = []
    for re in res:
        re2 = []
        for a,b in re:
            if a < 0:
                a = n+a
            if b < 0:
                b = m+b
            re2.extend((a+1,b+1))
        res2.append(re2)

    # console(res2)



    # console(grid)
    assert sum(sum(row) for row in grid) == 0

    for x1,y1,x2,y2,x3,y3 in res2:
        x1,y1,x2,y2,x3,y3 = x1-1,y1-1,x2-1,y2-1,x3-1,y3-1
        og[x1][y1] = 1 - og[x1][y1]
        og[x2][y2] = 1 - og[x2][y2]
        og[x3][y3] = 1 - og[x3][y3]
    assert sum(sum(row) for row in og) == 0

    assert len(res2) <= n*m

    return res2



def console(*args):  
    # print on terminal in different color
    print('\033[36m', *args, '\033[0m', file=sys.stderr)
    pass


# import random
# while True:
#     grid = [[int(random.random()) for _ in range(4)] for _ in range(4)]
#     solve_(grid,3,3)
#     # break



ONLINE_JUDGE = False

# if Codeforces environment
if os.path.exists('input.txt'):
    ONLINE_JUDGE = True

if ONLINE_JUDGE:
    sys.stdin = open("input.txt","r")
    sys.stdout = open("output.txt","w")

    def console(*args):
        pass


def solve(*args):
    # screen input
    if not ONLINE_JUDGE:
        console("----- solving ------")
        console(*args)
        console("----- ------- ------")
    return solve_(*args)


if True:
    # if memory is not a constraint
    inp = iter(sys.stdin.readlines())
    input = lambda: next(inp)
else:
    # if memory is a constraint
    input = sys.stdin.readline


for case_num in range(int(input())):
    # read line as a string
    # strr = input()

    # read line as an integer
    # k = int(input())
    
    # read one line and parse each word as a string
    # lst = input().split()

    # read one line and parse each word as an integer
    n,m = list(map(int,input().split()))

    # read matrix and parse as integers (after reading read nrows)
    # lst = list(map(int,input().split()))
    # nrows = lst[0]  # index containing information, please change
    grid = []
    for _ in range(n):
        grid.append(list(map(int,list(input().strip()))))

    res = solve(grid,n,m)  # please change
    
    # print result
    # Google - case number required
    # print("Case #{}: {}".format(case_num+1, res))

    # Codeforces - no case number required
    print(len(res))
    for r in res:
        print(*r)  # if printing a list