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:
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""

import os
import sys
from io import BytesIO, IOBase
import heapq as h 
from bisect import bisect_left, bisect_right
import time

from types import GeneratorType
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        import os
        self.os = os
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            self.os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        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")

from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string

#start_time = time.time()

def getInts():
    return [int(s) for s in input().split()]

def getInt():
    return int(input())

def getStrs():
    return [s for s in input().split()]

def getStr():
    return input()

def listStr():
    return list(input())

def getMat(n):
    return [getInts() for _ in range(n)]

MOD = 10**9+7

"""

"""
"""
Case where N and M are even: treat them as 2x2 blocks
Case where N is even, M is odd (or vv), treat them as 2x2 blocks until 3 from the end, then 2x3 blocks

Case where N and M are odd: treat them as 2x2 blocks, 2x3 blocks, 3x2 blocks, and one 3x3 block

Now I need to fix the 2 x 3 cases

"""



def solve():
    N, M = getInts()
    grid = []
    for n in range(N):
        grid.append([int(s) for s in input()])
    ans = 0
    ops = []
    def solve2(inds,flag=False,flag2=False):
        ret = []
        if not inds:
            return ret
        if len(inds) == 3:
            ret.append((inds[0][0],inds[0][1],inds[1][0],inds[1][1],inds[2][0],inds[2][1]))
            return ret
        if len(inds) == 1:
            neigh1 = (inds[0][0]+(1 if (inds[0][0] % 2  == 1 and inds[0][0] < N and not flag) or (flag and inds[0][0] < N)  else -1), inds[0][1]+(1 if (inds[0][1] % 2  == 1 and inds[0][1] < M and not flag2) or (flag2 and inds[0][1] < M) else -1))
            neigh2 = (inds[0][0]+(1 if (inds[0][0] % 2  == 1 and inds[0][0] < N and not flag) or (flag and inds[0][0] < N) else -1), inds[0][1])
            ret.append((inds[0][0],inds[0][1],neigh1[0],neigh1[1],neigh2[0],neigh2[1]))
            new_inds = [neigh1,neigh2]
            x = solve2(new_inds,flag,flag2)
            for el in x:
                ret.append(el)
            return ret
        if len(inds) == 4:
            ret.append((inds[0][0],inds[0][1],inds[1][0],inds[1][1],inds[2][0],inds[2][1]))
            new_inds = [(inds[3][0],inds[3][1])]
            x = solve2(new_inds,flag,flag2)
            for el in x:
                ret.append(el)
            return ret
        if len(inds) == 2:
            #choose 2 zeros and a one as first move
            if (inds[0][0]+inds[0][1]) % 2 == (inds[1][0]+inds[1][1]) % 2:
                neigh1 = (inds[0][0],inds[0][1]+(1 if (inds[0][1] % 2  == 1 and inds[0][1] < M and not flag2) or (flag2 and inds[0][1] < M) else -1))
                neigh2 = (inds[0][0]+(1 if (inds[0][0] % 2  == 1 and inds[0][0] < N and not flag) or (flag and inds[0][0] < N) else -1),inds[0][1])
            else:
                neigh1 = (inds[0][0],inds[0][1]+(1 if (inds[0][1] % 2  == 1 and inds[0][1] < M and not flag2) or (flag2 and inds[0][1] < M) else -1))
                neigh2 = (inds[0][0]+(1 if (inds[0][0] % 2  == 1 and inds[0][0] < N and not flag) or (flag and inds[0][0] < N) else -1),inds[0][1])
                if neigh1 == inds[1]:
                    neigh1 = (inds[0][0]+(1 if (inds[0][0] % 2  == 1 and inds[0][0] < N and not flag) or (flag and inds[0][0] < N) else -1),inds[0][1]+(1 if (inds[0][1] % 2  == 1 and inds[0][1] < M and not flag2) or (flag2 and inds[0][1] < M) else -1))
                else:
                    neigh2 = (inds[0][0]+(1 if (inds[0][0] % 2  == 1 and inds[0][0] < N and not flag) or (flag and inds[0][0] < N) else -1),inds[0][1]+(1 if (inds[0][1] % 2  == 1 and inds[0][1] < M and not flag2) or (flag2 and inds[0][1] < M) else -1))
            ret.append((inds[0][0],inds[0][1],neigh1[0],neigh1[1],neigh2[0],neigh2[1]))
            new_inds = [inds[1],neigh1,neigh2]
            x = solve2(new_inds,flag,flag2)
            for el in x:
                ret.append(el)
            return ret
    
    for n in range(N-1,1,-1):
        for m in range(M-1,1,-1):
            if grid[n][m] == 1:
                inds = (n+1,m+1,n,m+1,n,m)
                ops.append(inds)
                grid[n][m] ^= 1
                grid[n-1][m] ^= 1
                grid[n-1][m-1] ^= 1
        #grid[n][0] and grid[n][1]
        if grid[n][0] * grid[n][1] == 1:
            inds = (n+1,1,n+1,2,n,2)
            ops.append(inds)
            grid[n][0] = 0
            grid[n][1] = 0
            grid[n-1][1] ^= 1
        elif grid[n][0] == 1:
            inds = (n+1,1,n,1,n,2)
            ops.append(inds)
            grid[n][0] = 0
            grid[n-1][0] ^= 1
            grid[n-1][1] ^= 1
        elif grid[n][1] == 1:
            inds = (n+1,2,n,2,n,1)
            ops.append(inds)
            grid[n][1] = 0
            grid[n-1][1] ^= 1
            grid[n-1][0] ^= 1
    
    for m in range(M-1,1,-1):
        if grid[0][m]*grid[1][m] == 1:
            inds = (1,m+1,2,m+1,2,m)
            ops.append(inds)
            grid[0][m] = 0
            grid[1][m] = 0
            grid[1][m-1] ^= 1
        elif grid[0][m] == 1:
            inds = (1,m+1,1,m,2,m)
            ops.append(inds)
            grid[0][m] = 0
            grid[0][m-1] ^= 1
            grid[1][m-1] ^= 1
        elif grid[1][m] == 1:
            inds = (2,m+1,2,m,1,m)
            ops.append(inds)
            grid[1][m] = 0
            grid[1][m-1] ^= 1
            grid[0][m-1] ^= 1
    
    inds = []
    for i in range(2):
        for j in range(2):
            if grid[i][j] == 1:
                inds.append((i+1,j+1))
                grid[i][j] = 0
    tmp = solve2(inds)
    ans += len(tmp)
    for t in tmp:
        ops.append(t)
    
    print(len(ops))
    for op in ops:
        print(*op)
    return
    
for _ in range(getInt()):
    #print(solve())
    solve()

#print(time.time()-start_time)