// Slower C++ solution:
#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

const int maxn=1005;

int n,m,k,dx[]={-1,1,0,0},dy[]={0,0,1,-1},vis[maxn][maxn];

char map[maxn][maxn];

struct maze

{

    maze(int xx=0,int yy=0,int tt=0):x(xx),y(yy),time(tt){}

	int x,y,time;

};

maze bfs(int bx,int by,char ch)

{

	memset(vis,0,sizeof(vis));

	queue<maze> q;

	maze st(bx,by,0);

	q.push(st);vis[bx][by]=1;

	while(!q.empty())

	{

		st=q.front();q.pop();

		if(map[st.x][st.y]==ch)

		{

			return st;

		}

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

		{

			int tx=st.x+dx[i],ty=st.y+dy[i];

			if(tx<0||tx>=n||ty<0||ty>=m||map[tx][ty]=='X')//越界和?

			{

				continue;

			}

			maze tp(tx,ty,st.time+1);

			if(!vis[tx][ty])

			{

				vis[tx][ty]=1;

				q.push(tp);

			}

		}

	}

}

int main()

{



	while(~scanf("%d%d%d",&n,&m,&k))

	{

		int bx,by;

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

		{

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

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

			{

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

				{

					bx=i;by=j;

					

					map[i][j]='0'; //第0个起点

				}

			}

		}

		int ans=0;

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

		{

			maze tp=bfs(bx,by,map[bx][by]+1);

			ans+=tp.time;

			bx=tp.x;by=tp.y;

		}

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

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.4480434108501927x):
#include <iostream>

#include <stdlib.h>

#include <queue>

#include <cstring>

#define MAX_NUM 1010

using namespace std;

int H, W, N;

bool flag[MAX_NUM][MAX_NUM] = {false};                            //??是否已遍??

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



class Node {

  public:

    Node();

    Node(int x, int y, int step);

    ~Node();

    int x;

    int y;

    int step;

};



Node::Node() {

}



Node::Node(int x, int y, int step) {

    this->x = x;

    this->y = y;

    this->step = step;

}



Node::~Node() {

}



int BFS(char map[][MAX_NUM], int sx, int sy, int ex, int ey) {

    queue<Node> que;                                                  //第一次出?Memory Limit Exceeded,一定不能使用new来?行?建

    Node node(sx, sy, 0);

    que.push(node);

    flag[sx][sy] = true;                                              //将遍?的?点?置?true



    while(!que.empty()) {

        Node cur = que.front();

        que.pop();



        if(cur.x == ex && cur.y == ey) {

            return cur.step;



        } else {

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

                int k = cur.x + dir[i][0];

                int v = cur.y + dir[i][1];



                if(k >= 0 && k < H && v >= 0 && v < W && map[k][v] != 'X' && !flag[k][v]) { //将可到?并且尚未遍?的?点加入?列

                    Node temp(k, v, cur.step + 1);

                    que.push(temp);

                    flag[k][v] = true;

                }

            }

        }

    }



    return -1;

}



int main() {

    int sx, sy, arr[10][2];

    char map[MAX_NUM][MAX_NUM];



    if(cin >> H >> W >> N && H && W) {

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

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

                cin >> map[i][j];



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

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

                    sx = i;

                    sy = j;

                }



                if(isdigit(map[i][j])) {

                    int index = map[i][j] - '0';

                    arr[index][0] = i;

                    arr[index][1] = j;

                }

            }

        }



        arr[0][0] = sx;

        arr[0][1] = sy;

        int res = 0;



        for(int i = 0; i + 1 <= N; i++) {

            memset(flag, false, sizeof(flag));

            res += BFS(map, arr[i][0], arr[i][1], arr[i + 1][0], arr[i + 1][1]);

        }



        cout << res << endl;

    }



    return 0;

}