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

#include <queue>

#include <string.h>

using namespace std;

typedef pair<int, int> P;

int sx,sy,H,W,N;

char map1[1002][1002];

int step[1002][1002];

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

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

void find1(int g){

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

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

        if(map1[i][k] == 'S' || map1[i][k] == (g+48)){

        if(map1[i][k] == 'S') map1[i][k] ='.';

        sx = i; sy = k;

        return ;

    }

}



int bfs(int n)

{

    memset(step,-1,sizeof(step));

    queue <P> que;

    que.push(P (sx,sy));

    step[sx][sy] = 0;

    while(que.size()){

        int x = que.front().first, y = que.front().second;

        que.pop();

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

            int xn = x + dx[i], yn = y + dy[i];

            if(xn >= 0 && xn < W && yn >= 0 && yn < H && map1[xn][yn] != 'X' && step[xn][yn] == -1)

                {

                    step[xn][yn] = step[x][y] + 1;

                    if(map1[xn][yn] == (n+48))

                        {

                            return step[xn][yn];

                        }

                    que.push(P(xn, yn));

                }

        }

    }

return 0;

}



int main()

{

    cin >> W >> H >> N;

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

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

        cin >> map1[i][k];

    int step = 0;

    find1(0);

    step += bfs(1);

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

    {



        find1(i);

        step += bfs(i+1);

    }

    cout << step << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.4797972675581357x):
#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;

}