// 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: 2.5148719777185624x):
#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <vector>

#include <queue>



using namespace std;

struct as

{

    int x,y;

}pos[10]; //用来存储每次的起点和终点

struct node

{

    int x,y;

    int step;

};

bool vis[1010][1010];

char maze[1050][1050];

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

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

int H,W,N;



bool check(int x,int y)

{

    if (x<=0 || x>H) return 1;

    if (y<=0 || y>W) return 1;

    if (vis[x][y]) return 1;

    if (maze[x][y] =='X')return 1;

    return 0;

}

int bfs(int m,int n)

{

    memset(vis,0,sizeof vis);

    int sx = pos[m].x, sy = pos[m].y;

    int gx = pos[n].x, gy = pos[n].y;

   // printf("%d %d\n",sx,sy);

   // printf("%d %d\n",gx,gy);

    queue<node> Q;

    node a,next;

    a.step = 0;

    a.x = sx, a.y =sy;

    Q.push(a);

    while(!Q.empty())

    {

        a = Q.front();

        Q.pop();

        if (a.x == gx && a.y == gy)

            return a.step;

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

        {

            int nx = a.x + dx[i];

            int ny = a.y + dy[i];

            if(check(nx,ny))

                continue;

            else

            {

                vis[nx][ny] =1;

                next.x = nx,next.y = ny,next.step = a.step+1;

                Q.push(next);

            }

        }

    }

    return 999999;

}



int main()

{

   // freopen("in.txt","r",stdin);

    while(cin>>H>>W>>N)

    {

        int sx,sy;

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

        {

            for(int j = 1; j <= W;j++)

            {

                cin>>maze[i][j];

                if (maze[i][j] >='1' && maze[i][j]<='9')

                {

                    pos[maze[i][j]-'0'].x = i;

                    pos[maze[i][j]-'0'].y = j;

                }

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

                    sx=i,sy = j;

            }

        }



        pos[0].x = sx,pos[0].y = sy;

        int ans = 0;

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

            ans += bfs(i,i+1);

        cout<<ans<<endl;

    }

    return 0;

}