// Slower C++ solution:
/************************************************

* Author        :somniloquy

* Created Time  :2015/10/30 16:15:09

 ************************************************/



#include <cstdio>

#include <algorithm>

#include <iostream>

#include <cstring>

#include <queue>



using namespace std;



const int hw_max = 1000 + 5;

char graph[hw_max][hw_max];

int sign[hw_max][hw_max];

int h, w, n, ans;

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

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

struct node

{

	int x;

	int y;

	int time;

} start;



void BFS(node op, char ed)

{

	queue <node> process;

	process.push(op);

	sign[op.x][op.y] = 1;

	while(!process.empty())

	{

		node now = process.front();

		process.pop();

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

		{

			node next = now;

			next.x += dx[i];

			next.y += dy[i];

			next.time ++;

			if(next.x >= 0 && next.x < h && next.y >= 0 && next.y < w && graph[next.x][next.y] != 'X' && !sign[next.x][next.y])

			{

				if(graph[next.x][next.y] == ed)

				{

					ans += next.time;

					next.time = 0;

					start = next;

				       	return;

				}

				else

				{

					process.push(next);

					sign[next.x][next.y] = 1;

				}

			}

		}

	}

}



int main()

{

	while(~scanf("%d %d %d", & h, & w, & n))

	{

		getchar();

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

		{

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

			{

			       scanf("%c", & graph[i][j]);

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

			       {

					start.x = i;

					start.y = j;

					start.time = 0;

			       }

			}

			getchar();

		}

		ans = 0;

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

		{

			memset(sign, 0, sizeof(sign));

			BFS(start, '0' + i);

		}

		printf("%d\n", ans);

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.0762877787874965x):
#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;

}