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

#include<cstdio>

#include<queue>

#include<utility>

#include<cstring>

using namespace std;

char map[1005][1005];

bool vis[1005][1005];

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

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

int sx,sy;

int h,w;

int bfs(int n){

	memset(vis,false,sizeof(vis));

	queue<pair<pair<int,int>,int> >que;

	que.push(make_pair(make_pair(sx,sy),0));

	while(!que.empty()){

		pair<int,int>now=que.front().first;

		int cnt=que.front().second;

		que.pop();

		if(map[now.first][now.second]==n+48){

			sx=now.first;

			sy=now.second;

			return cnt;

		}

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

			int _x=now.first+dx[i];

			int _y=now.second+dy[i];

			if(_x<0||_y<0||_x>=h||_y>=w||map[_x][_y]=='X'||vis[_x][_y]){

				continue;

			}

			vis[_x][_y]=true;

			que.push(make_pair(make_pair(_x,_y),cnt+1));

		}

	}

}

int main(){

	int n;

	while(cin>>h>>w>>n){

		memset(map,0,sizeof(map));

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

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

		}

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

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

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

					sx=i;

					sy=j;

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

					break;

				}

			}

		}

		int count=0;

		for(int k=1;k<=n;k++){

			count+=bfs(k);

		}

		cout<<count<<endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4749965082800838x):
#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;

}