/* Solution to Pouring Honey by Mees
 *
 * Somewhat artificially takes very long at the end.
 *
 * Expected answer: TLE.
 *
 */
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>

using namespace std;

int main() {
    int H, m, n;
    cin >> H >> m >> n;
    vector<vector<int>> grid(m + 2, vector<int>(m + n + 2, 1));
    char c;
    for(int i = 1; i <= m; i++) {
        for(int j = 1 + i/2; j <= n + i/2; j++) {
            cin >> c;
            grid[i][j] = (c == '#');
        }
    }
    vector<pair<int, int>> offsets = {
        make_pair(1, 0),
        make_pair(-1, 0),
        make_pair(0, 1),
        make_pair(0, -1),
        make_pair(1, 1),
        make_pair(-1, -1)
    };

    vector<int> components;

    for(int i = 0; i < m + 2; i++) {
        for(int j = 0; j < m + n + 2; j++) {
            if(grid[i][j] == 0) {
                grid[i][j] = 1;
                int ct = 1;
                queue<pair<int, int>> q;
                q.push(make_pair(i, j));
                while(!q.empty()) {
                    auto p = q.front(); q.pop();
                    int curi = p.first;
                    int curj = p.second;
                    for(auto offset : offsets) {
                        int nbi = curi + offset.first;
                        int nbj = curj + offset.second;
                        if(grid[nbi][nbj] == 0) {
                            ct++;
                            grid[nbi][nbj] = 1;
                            q.push(make_pair(nbi, nbj));
                        }
                    }
                }
                components.push_back(ct);
            }
        }
    }

    int comp_ct = 0;

    while(H > 0) {
        int largest = -1;
        int idx = -1;
        for(int i = 0; i < (int)components.size(); i++) {
            if(components[i] > largest) {
                largest = components[i];
                idx = i;
            }
        }
        H -= largest;
        components[idx] = 0;
        comp_ct++;
    }

    cout << comp_ct << endl;
    return 0;
}