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

#include <cstdio>

#include <cstring>

#include <cmath>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <stack>

#include <utility>

#include <algorithm>

#include <sstream>

using namespace std;

typedef pair<int,int> P;



const int max_n=1000+10;

int h,w,n,sx,sy,dx[]={0,0,-1,1},dy[]={-1,1,0,0},d[max_n][max_n],ans;

char c[max_n][max_n];



void bfs(int x,int y,int m){

	if(m>n)return;

	queue<P> q;

	memset(d,0,sizeof(d));

	q.push(P(x,y));

	d[x][y]=1;

	while(!q.empty()){

		P p=q.front();

		q.pop();

		if(c[p.first][p.second]=='0'+m){

		    ans+=d[p.first][p.second]-1;

		    bfs(p.first,p.second,m+1);

		    return;

		}

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

			int tx=p.first+dx[i],ty=p.second+dy[i];

			if(tx>=0 && tx<h && ty>=0 && ty<w && c[tx][ty]!='X' && !d[tx][ty]){

				q.push(P(tx,ty));

				d[tx][ty]=d[p.first][p.second]+1;

			}

		}

	}

}



int main(){

    scanf("%d%d%d",&h,&w,&n);

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

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

            cin>>c[i][j];

            if(c[i][j]=='S'){sx=i;sy=j;}

        }

    bfs(sx,sy,1);

    cout<<ans<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.5140430811180385x):
#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;

}