This is the hard 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 nm 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 ≤ nm) — 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 to 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  
    
"""
                            pppppppppppppppppppp
                         ppppp  ppppppppppppppppppp
                      ppppppp    ppppppppppppppppppppp
                      pppppppp  pppppppppppppppppppppp
                      pppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
       ppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppp
      pppppppppppppppppppppppppppppppppppppppppppppppp  ppppppppppppppppppppp
     ppppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppppp
    ppppppppppppppppppppppppppppppppppppppppppppppp    pppppppppppppppppppppppp
   pppppppppppppppppppppppppppppppppppppppppppppp     pppppppppppppppppppppppppp
  ppppppppppppppppppppppppppppppppppppppppppppp      pppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppppppp               pppppppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppp     pppppppppppppppppppppppppppppppppppppppppppppp
  ppppppppppppppppppppppppppp    pppppppppppppppppppppppppppppppppppppppppppppppp
    pppppppppppppppppppppppp  pppppppppppppppppppppppppppppppppppppppppppppppppp
     ppppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppppp
      pppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppp
       ppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
                              pppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppp  pppppppp
                              ppppppppppppppppppppp    ppppppp
                                 ppppppppppppppppppp  ppppp
                                    pppppppppppppppppppp
"""


import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")

mod = pow(10, 9) + 7
# mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var)+"\n")
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


def perform(z, o):
    answer.append([])
    for first, second in nei:
        if z and mat[first][second] == '0':
            answer[-1].append((first + 1, second + 1))
            z -= 1
            mat[first][second] = '1'
        elif o and mat[first][second] == '1':
            answer[-1].append((first + 1, second + 1))
            o -= 1
            mat[first][second] = '0'


for _ in range(int(data())):
    n, m = sp()
    mat = [list(data()) for __ in range(n)]
    answer = []
    if n & 1 or m & 1:
        # Change last row
        i = 0
        j = m - 1
        while i < n:
            if mat[i][j] == mat[i + 1][j] == '0':
                i += (1, 2)[i != n - 3]
                continue
            answer.append([])
            if mat[i][j] == '1':
                mat[i][j] = '0'
                answer[-1].append([i + 1, j + 1])
            if mat[i + 1][j] == '1':
                mat[i + 1][j] = '0'
                answer[-1].append([i + 2, j + 1])
            if len(answer[-1]) == 1:
                answer[-1].append([i + 1, j])
                answer[-1].append([i + 2, j])
                mat[i][j - 1] = ('0', '1')[mat[i][j - 1] == '0']
                mat[i + 1][j - 1] = ('0', '1')[mat[i + 1][j - 1] == '0']
            elif len(answer[-1]) == 2:
                answer[-1].append([i + 2, j])
                mat[i + 1][j - 1] = ('0', '1')[mat[i + 1][j - 1] == '0']
            i += (1, 2)[i != n - 3]
        # Change last column
        j = 0
        i = n - 1
        while j < m:
            if mat[i][j] == mat[i][j + 1] == '0':
                j += (1, 2)[j != m - 3]
                continue
            answer.append([])
            if mat[i][j] == '1':
                mat[i][j] = '0'
                answer[-1].append([i + 1, j + 1])
            if mat[i][j + 1] == '1':
                mat[i][j + 1] = '0'
                answer[-1].append([i + 1, j + 2])
            if len(answer[-1]) == 1:
                answer[-1].append([i, j + 1])
                answer[-1].append([i, j + 2])
                mat[i - 1][j] = ('0', '1')[mat[i - 1][j] == '0']
                mat[i - 1][j + 1] = ('0', '1')[mat[i - 1][j + 1] == '0']
            elif len(answer[-1]) == 2:
                answer[-1].append([i, j + 2])
                mat[i - 1][j + 1] = ('0', '1')[mat[i - 1][j + 1] == '0']
            j += (1, 2)[j != m - 3]
    i = 0
    while i < n:
        j = 0
        while j < m:
            nei = [[i, j], [i, j + 1], [i + 1, j], [i + 1, j + 1]]
            j += (1, 2)[j != m - 3]
            cnt = 0
            for u, v in nei:
                cnt += int(mat[u][v])
            # print(cnt)
            if cnt == 0:
                continue
            if cnt == 1:
                perform(2, 1)
                perform(2, 1)
                perform(0, 3)
            elif cnt == 2:
                perform(2, 1)
                perform(0, 3)
            elif cnt == 3:
                perform(0, 3)
            else:
                perform(0, 3)
                perform(2, 1)
                perform(2, 1)
                perform(0, 3)
        i += (1, 2)[i != n - 3]
    # if len(answer) > n * m:
    #     print(_, n, m, len(answer))
    #     print("FUCKED")
    #     exit()
    # print(mat)
    out(len(answer))
    for i in answer:
        for a, b in i:
            outa(a, b, end=" ")
        outa()
