Codeforces Round ＃684 (Div. 1)

A1. Binary Table (Easy Version)

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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×mn×m. This table consists of symbols 00 and 11.

You can make such operation: select 33 different cells that belong to one 2×22×2 square and change the symbols in these cells (change 00 to 11 and 11 to 00).

Your task is to make all symbols in the table equal to 00. You are allowed to make at most 3nm3nm 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 tt (1≤t≤50001≤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 nn, mm (2≤n,m≤1002≤n,m≤100).

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

It is guaranteed that the sum of nmnm for all test cases does not exceed 2000020000.

Output

For each test case print the integer kk (0≤k≤3nm0≤k≤3nm) — the number of operations.

In the each of the next kk lines print 66 integers x1,y1,x2,y2,x3,y3x1,y1,x2,y2,x3,y3 (1≤x1,x2,x3≤n,1≤y1,y2,y3≤m1≤x1,x2,x3≤n,1≤y1,y2,y3≤m) describing the next operation. This operation will be made with three cells (x1,y1)(x1,y1), (x2,y2)(x2,y2), (x3,y3)(x3,y3). These three cells should be different. These three cells should belong into some 2×22×2 square.

Example

input

Copy

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

Copy

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)(1,1), (2,1)(2,1), (2,2)(2,2). After that, all symbols will be equal to 00.

In the second test case:

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

  011
  001
  000
  

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

  000
  000
  000
  

In the fifth test case:

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

  010
  110
  

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

  000
  000

• ============================================================================================================================================

之前出过类似的题目，这类题除了针对一个一个进行填补或归位没有任何好方法，突破点就在限制的次数上，3*n*m 就意味着一个点构成的2*2方阵，要想单独改变一个而不去改变其余三个，需要进行三次操作。演草可以退出四个位置的1的改变方法，然后放在图上进行分类讨论即可

# include<iostream>
using namespace std;
char s[110][110];

int main ()
{

    int t;
    cin>>t;

    while(t--)
    {
        int n,m;
        cin>>n>>m;

        int ans=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                cin>>s[i][j];
                
                ans+=(s[i][j]=='1');

            }
        }
        cout<<ans*3<<endl;

        for(int i=1;i<=n-1;i++)
        {
            for(int j=1;j<=m-1;j++)
            {
                if(s[i][j]=='1')
                {
                    cout<<i<<" "<<j<<" "<<i+1<<" "<<j<<" "<<i+1<<" "<<j+1<<endl;

                    cout<<i<<" "<<j<<" "<<i<<" "<<j+1<<" "<<i+1<<" "<<j<<endl;

                    cout<<i<<" "<<j<<" "<<i<<" "<<j+1<<" "<<i+1<<" "<<j+1<<endl;

                }
            }
        }

        for(int i=1;i<=n-1;i++)
        {
            if(s[i][m]=='1')
            {
                int j=m;

                cout<<i<<" "<<j<<" "<<i+1<<" "<<j-1<<" "<<i+1<<" "<<j<<endl;

                cout<<i<<" "<<j-1<<" "<<i<<" "<<j<<" "<<i+1<<" "<<j-1<<endl;

                cout<<i<<" "<<j<<" "<<i<<" "<<j-1<<" "<<i+1<<" "<<j<<endl;
            }
        }

       if(s[n][1]=='1')
       {
           int i=n,j=1;

           cout<<i<<" "<<j<<" "<<i-1<<" "<<j<<" "<<i-1<<" "<<j+1<<endl;

           cout<<i<<" "<<j<<" "<<i<<" "<<j+1<<" "<<i-1<<" "<<j<<endl;

           cout<<i<<" "<<j<<" "<<i<<" "<<j+1<<" "<<i-1<<" "<<j+1<<endl;
       }

       for(int j=2;j<=m;j++)
       {
           if(s[n][j]=='1')
           {
               int i=n;

               cout<<i<<" "<<j<<" "<<i-1<<" "<<j-1<<" "<<i<<" "<<j-1<<endl;

               cout<<i<<" "<<j<<" "<<i-1<<" "<<j<<" "<<i-1<<" "<<j-1<<endl;

               cout<<i<<" "<<j<<" "<<i-1<<" "<<j<<" "<<i<<" "<<j-1<<endl;
           }
       }
    }
    return 0;
}