The Pilots Brothers' refrigerator

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

Sample Input

-+--
----
----
-+--

Sample Output

6
1 1
1 3
1 4
4 1
4 3
4 4

#include<iostream>
#include<cstring>
#include<string>
#include<queue>
using namespace std;
int wei[20]= { 63624, 62532, 61986,61713, 36744,20292,12066,7953,35064, 17652,8946,4593,34959, 17487, 8751, 4383};//16种状态
bool vis[65536+10];
int dis[65536+10];

int num;

struct node
{
    int t;
    int step;
};
void bfs()
{
    queue<node>Q;
    node now,next;
    now.t=num;
    now.step=0;
    Q.push(now);
    while(!Q.empty())
    {
        now=Q.front();
        Q.pop();
        for(int i=0; i<16; i++)
        {
            next.t=now.t^wei[i];//取异或变状态
            if(!vis[next.t])//判断当前状态是否出现
            {
                dis[next.t]=i;
                next.step=now.step+1;
                vis[next.t]=true;
                Q.push(next);
            }
            if(next.t==0)//判断是否达到目标状态
            {
                cout<<now.step+1<<endl;
                int j=next.t;
                while(j!=num)//输出坐标位置
                {
                    cout<<dis[j]/4+1<<" "<<dis[j]%4+1<<endl;
                    j=j^wei[dis[j]];
                }
                return ;
            }
        }
    }
    return ;
}

int main()
{
    char str;
    memset(vis,false,sizeof(vis));
    num=0;
    for(int i=0; i<4; i++)
    {
        for(int j=0; j<4; j++)//状态压缩，变为具体的数
        {
            cin>>str;
            if(str=='+')
                num=(num<<1)+1;
            else num=(num<<1);
        }
    }
    //cout<<num<<endl;
    if(num==0)
    {
        cout<<0<<endl;
        return 0;
    }
    else
    {
        vis[num]=true;
        bfs();
    }
    return 0;
}