// 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: 4.675358838459754x):
#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <queue>

using namespace std;



const int N = 1000;

const int INF = 0x3f3f3f3f;



int n, m, f;

char map[N][N+1];



bool legal(int x, int y, int k)

{

    return x >= 0 && x < n && y >= 0 && y < m

        && map[x][y] != 'X';

}



void input()

{

    cin >> n >> m >> f;

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

        scanf("%s", map[i]);

    }

}



int BFS(int k)

{

    int i, j;

    int d[N][N];

    typedef pair<int, int> P;

    queue <P> q;

    for (i = 0; i < n; i ++) {

        for (j = 0; j < m; j ++) {

            d[i][j] = INF;

            if ((k == 0 && map[i][j] == 'S') || map[i][j] == '0'+k) {

                q.push(P(i, j));

                d[i][j] = 0;

                map[i][j] = '.';

            }

        }

    }



    while( q.size() ) {

        P p = q.front();

        q.pop();

        int x = p.first;

        int y = p.second;



        if (map[x][y] == INF)

            return d[x][y];



        int pos[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

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

            int nx = x + pos[i][0];

            int ny = y + pos[i][1];

            if (legal(nx, ny, k) && d[nx][ny] == INF) {

                q.push(P(nx, ny));

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

                if (map[nx][ny] == '0'+k+1) {

                    return d[nx][ny];

                }

            }

        }

    }



    return INF;

}



void solve()

{

    int res = 0;

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

        res += BFS(i);

    }

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

}



int main(void)

{

    input();



    solve();



    return 0;

}