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

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<string>

#include<cstdlib>

#include<stack>

#include<queue>

#include <map>

#include<cmath>

#include<vector>

#define MAX_N 1000000

#define INF 0x3f3f3f3f

#define SIZE 1000

using namespace std;







struct note

{

    int x, y;

}start, ende;

typedef pair<int, int> P;

int H, W, N;

char a[1003][1003];

int sum;

int d[1003][1003];

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

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

void bfs(note h, note m)

{

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

    queue<P> que;

    que.push(P(h.x, h.y));

    d[h.x][h.y] = 0;

    while (!que.empty())

    {

        P p = que.front();

        que.pop();

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

        {

            int xx = p.first + dx[i];

            int yy = p.second + dy[i];

            if (xx >= 0 && xx < H && yy >= 0 && yy < W && a[xx][yy] != 'X' && d[xx][yy] == -1)

            {

                d[xx][yy] = d[p.first][p.second] + 1;

                que.push(P(xx, yy));

            }

        }

    }

    sum += d[m.x][m.y];

    start = ende;

}

void solve()

{

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

    {

        for (int j = 0; j < W; j++)

        {

            if (a[i][j] == 'S')

            {

                start.x = i;

                start.y = j;

                i = H;

                j = W;

            }

        }

    }

    for (int k = 0; k < N; k++)

    {

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

        {

            for (int j = 0; j < W; j++)

            {

                if (a[i][j] == k + 49)

                {

                    ende.x = i;

                    ende.y = j;

                    bfs(start, ende);

                    i = H;

                    j = W;

                }

            }

        }

    }

    cout << sum << endl;

}

int main()

{

    cin >> H >> W >> N;

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

    {

        cin >> a[i];

    }

    sum = 0;

    solve();

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.0690799874411585x):
#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;

}