// Slower C++ solution:
#include <queue>

#include <cstring>

#include <iostream>

using namespace std;



const int N = 1111;

const int cx[] = {0, 0, 1, -1}, cy[] = {1, -1, 0, 0};

char maze[N][N];

int step[N][N];

int h, w, n, sx, sy, ans;



int bfs(int cheese){

    memset(step, -1, sizeof(step));

    queue<pair<int, int> > que;

    que.push(make_pair(sx, sy));

    step[sx][sy] = 0;

    while(!que.empty()){

        int x = que.front().first, y = que.front().second;

        que.pop();

        for(int i = 0; i < 4; ++i){

            int nx = x + cx[i], ny = y + cy[i];

            if(nx < 0 || h <= nx || ny < 0 || w <= ny || maze[nx][ny] == 'X' || step[nx][ny] != -1) continue;

            step[nx][ny] = step[x][y] + 1;

            if(maze[nx][ny] - '0' == cheese){

                sx = nx;

                sy = ny;

                return step[nx][ny];

            }

            que.push(make_pair(nx, ny));

        }

    }

    return 0;

}



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    while(cin >> h >> w >> n){

        for(int i = 0; i < h; ++i)

            for(int j = 0; j < w; ++j){

                cin >> maze[i][j];

                if(maze[i][j] == 'S'){

                    sx = i;

                    sy = j;

                }

            }

        ans = 0;

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

            ans += bfs(i);

        cout << ans << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 6.0514569804387195x):
#include <iostream>

#include <stdio.h>

#include <deque>



using namespace std;



const int INF = 0x3f3f3f3f;

typedef pair<int, int> P;

int h, w, n;

int sx, sy;

int gx, gy;

char cmap[1000][1000];

int goalSum;

int goal[10];

int d[1000][1000];

int dx[4] = {-1, 0, 1, 0};

int dy[4] = {0, -1, 0, 1};



int bfs() {

    int res = 0;

    deque<P> que;

    for (int ci = 0; ci < n; ci++) {  //要?行n次循?，?找n次?点，?次?数?入goal[i], 最后累加起来是res 

        for (int i = 0; i < h; i++) {

            for (int j = 0; j < w; j++) {

                d[i][j] = INF;

            }

        }

        

        que.push_back(P(sx, sy));

        d[sx][sy] = 0;

        

        while(que.size()) {            

            int x = que.front().first;

            int y = que.front().second;

                        

            if ((int)cmap[x][y] == 49 + ci) {

                goal[ci] = d[x][y];

                sx = x;

                sy = y;

                break;

            }

            

            for (int i = 0; i < 4; i++) {

                int nx = x + dx[i];

                int ny = y + dy[i];

                if (nx >=0 && nx < h && ny >= 0 && ny < w && cmap[nx][ny] != 'X' && d[nx][ny] == INF) {

                    que.push_back(P(nx, ny));

                    d[nx][ny] = d[x][y] + 1;

                }

            }    

            que.pop_front();        

        }

        que.clear(); 

        res += goal[ci];

    }

    return res;

}



int main() {

    cin >> h >> w >> n;

    for (int i = 0; i < h; i++) {

        for (int j = 0; j < w; j++) {

            cin >> cmap[i][j];

            if (cmap[i][j] == 'S') {

                sx = i;

                sy = j;

            }

        }

    }

    goalSum = bfs();

    cout << goalSum << endl;

    return 0;

}