// Original C++ solution (slower):
#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;

}




// Optimized solution to the same problem (faster):
#include <iostream>

#include <queue>

using namespace std;

const int INF = 100000000;

typedef pair<int,int> Point;

Point AllPoint[1000];

int W,H;

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

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

char map[1000][1000];

int deep[1000][1000];

int N;

void BFS(Point Start,Point Over){

    queue<Point> que;

    int FirstDeep = deep[Start.first][Start.second];

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

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

            deep[i][j] = INF;

        }

    }

    que.push(Start);

    deep[Start.first][Start.second] = FirstDeep;

    while (que.size()){

        Point p = que.front();

        que.pop();

        if (p.first== Over.first&&p.second == Over.second)

            break;

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

            int ix = p.first+x[i];

            int iy = p.second+y[i];

            if (ix>=0&&ix<=H

                &&iy>=0&&iy<=W

                &&deep[ix][iy] == INF

                &&map[ix][iy]!='X'){

                que.push(Point(ix,iy));

                deep[ix][iy] = deep[p.first][p.second]+1;

            }

        }

    }

}







int main(){

    cin >> H >> W;

    cin >> N;

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

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

            cin >> map[i][j];

            if ((int)map[i][j]-48>=1&&(int)map[i][j]-48<=N){

                AllPoint[(int)map[i][j]-48].first = i;

                AllPoint[(int)map[i][j]-48].second = j;



            }

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

                AllPoint[0].first = i;

                AllPoint[0].second = j;

            }

        }

    }

    deep[AllPoint[0].first][AllPoint[0].second] = 0;

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

        BFS(AllPoint[i],AllPoint[i+1]);

    }

    cout << deep[AllPoint[N].first][AllPoint[N].second]<<endl;



    return 0;

}

//#....

//S.##.

//###G.

//#####

//#####



//#S######.#

//......#..#

//.#.##.##.#

//.#........

//##.##.####

//....#....#

//.#######.#

//....#.....

//.####.###.

//....#...G#






// Fastest known solution for this problem: