/*

     题意：4*4的矩形，改变任意点，把所有'+'变成'-',，每一次同行同列的都会反转，求最小步数，并打印方案

     DFS：把'+'记为1， '-'记为0

     1. 从(1, 1)走到(4, 4)，每一点DFS两次(改点反转或不反转)；used记录反转的位置

         详细解释：http://poj.org/showmessage?message_id=346723

     2. 比较巧妙的解法：抓住'+'位置反转，'-'位置相对不反转的特点，从状态上考虑

         详细解释：http://poj.org/showmessage?message_id=156561

     3. 枚举步骤数(1~16)，暴力解法，耗时大

         详细解释：http://poj.org/showmessage?message_id=343281

     4. 网上还有其他解法：高斯消元，BFS，+位运算等等

     注意：反转时十字形中心位置多反转了两次，要再反转一次

     我还是DFS写不出来。。。

 */

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <map>

 #include <queue>

 #include <vector>

 using namespace std;

 const int MAXN = 1e6 + ;

 const int INF = 0x3f3f3f3f;

 int a[][];

 int used[][];

 bool ok(void)

 {

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

     {

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

             if (a[i][j] == )    return false;

     }

     return true;

 }

 void change(int x, int y)

 {

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

     {

         a[x][i] = a[x][i] ?  : ;

         a[i][y] = a[i][y] ?  : ;

     }

     a[x][y] = a[x][y] ?  : ;

     used[x][y] = used[x][y] ?  : ;

 }

 bool DFS(int x, int y)

 {

     if (x ==  && y == )

     {

         if (ok ())    return true;

         change (x, y);

         if (ok ())    return true;

         change (x, y);

         return false;

     }

     int nx, ny;

     if (y == )        nx = x + , ny = ;

     else    nx = x, ny = y + ;

     if (DFS (nx, ny))    return true;

     change (x, y);

     if (DFS (nx, ny))    return true;

     change (x, y);

     return false;

 }

 void work(void)

 {

     if (DFS (, ))

     {

         int ans = ;

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

         {

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

             {

                 if (used[i][j])        ans++;

             }

         }

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

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

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

                 if (used[i][j])    printf ("%d %d\n", i, j);

     }

     else    printf ("WA\n");

 }

 int main(void)        //POJ 2965 The Pilots Brothers' refrigerator

 {

     //freopen ("B.in", "r", stdin);

     char ch;

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

     {

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

         {

             scanf ("%c", &ch);

             a[i][j] = (ch == '-') ?  : ;

             //printf ("%d ", a[i][j]);

         }

         getchar ();    //puts ("");

     }

     work ();

     return ;

 }

 /*

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <map>

 #include <queue>

 #include <vector>

 using namespace std;

 const int MAXN = 1e6 + 10;

 const int INF = 0x3f3f3f3f;

 bool con[5][5];

 int a[5][5];

 struct NODE

 {

     int x, y;

 }node[MAXN];

 void work(void)

 {

     for (int i=1; i<=4; ++i)

     {

         for (int j=1; j<=4; ++j)

         {

             if (a[i][j] == 1)

             {

                 con[i][j] = !con[i][j];

                 for (int k=1; k<=4; ++k)

                 {

                     con[i][k] = !con[i][k];

                     con[k][j] = !con[k][j];

                 }

             }

         }

     }

     int ans = 0;

     for (int i=1; i<=4; ++i)

     {

         for (int j=1; j<=4; ++j)

         {

             if (con[i][j] == true)

             {

                 ans++;    node[ans].x = i;    node[ans].y = j;

             }

         }

     }

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

     for (int i=1; i<=ans; ++i)

         printf ("%d %d\n", node[i].x, node[i].y);

 }

 int main(void)        //POJ 2965 The Pilots Brothers' refrigerator

 {

     //freopen ("B.in", "r", stdin);

     char ch;

     for (int i=1; i<=4; ++i)

     {

         for (int j=1; j<=4; ++j)

         {

             scanf ("%c", &ch);

             a[i][j] = (ch == '-') ? 0 : 1;

         }

         getchar ();

     }

     work ();

     return 0;

 }

 */

 /*

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <map>

 #include <queue>

 #include <vector>

 using namespace std;

 const int MAXN = 1e6 + 10;

 const int INF = 0x3f3f3f3f;

 int a[5][5];

 struct NODE

 {

     int x, y;

 }node[MAXN];

 int k;

 bool flag;

 bool ok(void)

 {

     for (int i=1; i<=4; ++i)

     {

         for (int j=1; j<=4; ++j)

             if (a[i][j] == 1)    return false;

     }

     return true;

 }

 void change(int x, int y)

 {

     for (int i=1; i<=4; ++i)

     {

         a[x][i] = !a[x][i];

         a[i][y] = !a[i][y];

     }

     a[x][y] = !a[x][y];

 }

 void DFS(int x, int y, int num, int cnt, int kk)

 {

     if (num == cnt)

     {

         flag = ok ();

         k = kk;

         return ;

     }

     for (int i=x; i<=4; ++i)

     {

         int j;

         if (i == x)    j = y + 1;

         else    j = 1;

         for (; j<=4; ++j)

         {

             node[kk].x = i;

             node[kk].y = j;

             change (i, j);

             DFS (i, j, num+1, cnt, kk+1);

             if (flag)    return ;

             change (i, j);

         }

     }

 }

 void work(void)

 {

     int cnt;

     for (cnt=1; cnt<=16; ++cnt)

     {

         flag = false;    k = 0;

         DFS (1, 0, 0, cnt, 1);

         if (flag)    break;

     }

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

     for (int i=1; i<=k - 1; ++i)

         printf ("%d %d\n", node[i].x, node[i].y);

 }

 int main(void)        //POJ 2965 The Pilots Brothers' refrigerator

 {

     //freopen ("B.in", "r", stdin);

     char ch;

     for (int i=1; i<=4; ++i)

     {

         for (int j=1; j<=4; ++j)

         {

             scanf ("%c", &ch);

             a[i][j] = (ch == '-') ? 0 : 1;

         }

         getchar ();

     }

     work ();

     return 0;

 }

 */