/* Topcoder SRM 700, D2, 500-Pointer (XMarksTheSpot)
 * James S. Plank
 * Tue Oct 25 15:18:18 EDT 2016
 */

#include <string>
#include <vector>
#include <list>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;

class XMarksTheSpot
{
    public:
        int countArea(vector <string> board);
};

int XMarksTheSpot::countArea(vector <string> B)
{
    int i, j;
    vector <int> QR, QC;                          // Row/columns of the question marks.
    int minrow;                                   // Minimum row for the monuments ('O')
    int maxrow;                                   // Maximum row for the monuments ('O')
    int mincol;                                   // Minimum col for the monuments ('O')
    int maxcol;                                   // Maximum col for the monuments ('O')
    int rv;                                       // Return value.

    /* During the enumeration, you need to calculate the min/max row/column when
       you consider each subset.  I use r, R, c and C to do that calculation. */

    int r, R, c, C;

    /* Sentinelize the min/max row/col. */

    minrow = B.size();
    maxrow = -1;
    mincol = B[0].size();
    maxcol = -1;

    /* Here's where we create QR and QC, and also calculate the min/max row/col */

    for (i = 0; i < B.size(); i++) {
        for (j = 0; j < B[i].size(); j++) {
            if (B[i][j] == '?') {
                QR.push_back(i);
                QC.push_back(j);
            }
            else if (B[i][j] == 'O') {
                if (i < minrow) minrow = i;
                if (i > maxrow) maxrow = i;
                if (j < mincol) mincol = j;
                if (j > maxcol) maxcol = j;
            }
        }
    }

    /* The constraints tell you that there is always at least one 'O', so you don't have
       to check whether minrow still equals B.size() or maxrow still equals -1. */

    rv = 0;
    printf("<table>\n");

    /* Do the power set enumeration of all subsets of question marks */

    for (i = 0; i < (1 << QR.size()); i++) {

        /* You need to calculate the min/max row/column which includes the subset
           of question marks.  Start with the min/max row/column of the known monuments. */

        r = minrow;
        R = maxrow;
        c = mincol;
        C = maxcol;

        /* For each subset, run through and modify r/R/c/C if the question mark changes it. */

        for (j = 0; j < QR.size(); j++) {
            if (i & (1 << j)) {
                if (QR[j] < r) r = QR[j];
                if (QR[j] > R) R = QR[j];
                if (QC[j] < c) c = QC[j];
                if (QC[j] > C) C = QC[j];
            }
        }

        /* Add the rectangle size to the return value. */

        rv += ( (R-r+1)*(C-c+1) );

        /* I'm printing the HTML of that table. */

        printf("<tr>\n");
        printf("<td align=center>{");
        for (j = 0; j < QR.size(); j++) {
            if (i & (1 << j))  printf(" %d", j);
        }
        printf(" }</td>\n");
        printf("<td align=center>0x%x</td>\n", i);
        printf("<td align=center>%d</td>\n", r);
        printf("<td align=center>%d</td>\n", R);
        printf("<td align=center>%d</td>\n", c);
        printf("<td align=center>%d</td>\n", C);
        printf("<td align=center>%d</td>\n", ( (R-r+1)*(C-c+1) ));
        printf("<td align=center>%d</td>\n", rv);
        printf("</tr>\n");
    }

    printf("</table>\n");

    /* Return the final value. */

    return rv;
}