//Solution by Zhusupov Nurlan
#include <iostream>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <algorithm>
#include <set>
#include <vector>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <ctime>

using namespace std;

typedef long long LL;
typedef map<string , int> MSI;
typedef vector<int> VI;

#define pb(x) push_back(x)
#define sqr(x) ((x)*(x))
#define F first
#define S second
#define SZ(t) ((int)t.size())
#define len(t) ((int)t.length())
#define base 10
#define fname "coins."
#define sz 1000*1000
#define EPS (1e-8)
#define INF ((int)1e9 + 9)
#define write(xx) printf("%d" , xx);
#define readln(xx) getline(cin , xx)
#define read(xx) scanf("%d" , &xx)
#define for(xx1 , yy1 , zz1) for(int zz1 = xx1 ; zz1 <= yy1 ; zz1++)

const double PI  = acos(-1.0);

int s, t, k, n, m, ans;
char a[100][100];

int main(){
    freopen(fname"in", "r", stdin);
    freopen(fname"out", "w", stdout);

    cin >> n >> m >> k;

    for (1 , n , i)
        for (1 , m , j)
        {
            cin >> a[i][j];
            s += (a[i][j] == 'o');
        }
    ans = INF;

    for (1 , n , i1)
        for (i1 , n , i2)
            for (1 , m , j1)
                for (j1 , m , j2)
                    {
                        t = 0;
                        for (i1 , i2 , i)
                            for (j1 , j2 , j)
                                t += (a[i][j] == 'o');
                        if (t == k)
                        {
                            ans = min(ans, (i1 + j1 - 2) * 2 + n - i2 + m - j2);
                            ans = min(ans, (n - i2 + m - j2) * 2 + i1 + j1 - 2);
                            ans = min(ans, (j1 - 1 + n - i2) * 2 + i1 - 1 + m - j2);
                            ans = min(ans, (i1 - 1 + m - j2) * 2 + j1 - 1 + n - i2);
                        }
                    }
    if (ans == INF)
        cout << -1;
    else
        cout << ans;
}