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.

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.

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.

 #include<stdio.h>

 #include<string.h>

 int main() {

     int in[];

     int tmp[];

     int length=;

     int size=length*length;

     char temp;

     //将输入变成0,1，开的为1，关的为0，放在数组in中

     for(int i=;i<size;i++){

         scanf("%c",&temp);

         if(temp=='\n')

             scanf("%c",&temp);

         if(temp=='+')

             in[i]=;

         else

             in[i]=;

     }

 //    for(int i=0;i<size;i++){

 //        printf("%d \n",in[i]);

 //    }

     //com数组存放的是开关的组合情况

     int com[];

     for(int i=;i<size+;i++){

         for(int j=;j<i;j++){

             com[j]=j;

         }

         int end=;

         while(){

             memcpy(tmp,in,size*sizeof(int));

             //按照搜索到的组合情况对冰箱进行开关

             for(int j=;j<i;j++){

             int row=com[j]/;

             int col=com[j]%;

             for(int k=;k<length;k++){

                 tmp[row*length+k]=-tmp[row*length+k];

                 tmp[k*length+col]=-tmp[k*length+col];

             }

             tmp[row*length+col]=-tmp[row*length+col];

             }

             int zero=;

             for(int j=;j<size;j++){

                 if(tmp[j]==){

                     zero+=;

                 }

             }

 //            for(int k=0;k<4;k++){

 //                printf("%d\t",com[k]);

 //            }

 //            printf("\n");

     //        printf("sero is %d\n",zero);

 //            zero=size;

             //判断是否找到答案

             if(zero==size){

                 end=;

                 break;

             }

             //判断开关情况是i个的组合是否已经全部搜索完毕，是则搜索i+1个情况

             if(com[]==size-i)

                 break;

             //搜索下一个开关为i个的组合情况

             for(int j=i-;j>=;j--){

                 if(com[j]!=size-(i-j)){

                     com[j]++;

                     for(int k=j+;k<i;k++)

                         com[k]=com[k-]+;

                     break;

                 }

             }

         }

         //若找到结果，输出结果

         if(end==){

             printf("%d\n",i);

             for(int j=;j<i;j++){

                 printf("%d %d\n",com[j]/+,com[j]%+);

             }

         }

     }

     return ;

 }