// Original C++ solution (slower):
#include <iostream>

#include <cctype>

#include <queue>

#include <string.h>

using namespace std;

const int DIRECTSIZE = 4;

struct _direct {

    int drow;

    int dcol;

} direct[DIRECTSIZE] = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};



const int N = 1000;

const int N2 = 10;

char grid[N][N], visited[N][N];

int h, w, n;

struct _factory {

    int row, col;

} factory[N2];



struct _node {

    int row, col, level;

};



int bfs(int n, char goal)

{

    memset(visited, 0, sizeof(visited));



    _node start={factory[n-1].row, factory[n-1].col, 0}, temp;



    visited[start.row][start.col] = 1;



    queue<_node> q;

    q.push(start);



    while(!q.empty()) {

        _node front = q.front();

        q.pop();



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

            int nextrow = front.row + direct[i].drow;

            int nextcol = front.col + direct[i].dcol;

            if( 0 <= nextrow && nextrow < h && 0 <= nextcol && nextcol < w) {

                if(grid[nextrow][nextcol] == goal) {

                    return front.level + 1;

                } else if(visited[nextrow][nextcol] == 0 && grid[nextrow][nextcol] != 'X') {

                    visited[nextrow][nextcol] = 1;

                    temp.row = nextrow;

                    temp.col = nextcol;

                    temp.level = front.level + 1;

                    q.push(temp);

                }

            }

        }

    }



    return 0;

}



int main()

{

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

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

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

                cin >> grid[i][j];



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

                    factory[0].row = i;

                    factory[0].col = j;

                } else if(isdigit(grid[i][j])) {

                    factory[grid[i][j] - '0'].row = i;

                    factory[grid[i][j] - '0'].col = j;

                }

            }



        int minstep = 0;

        char goal = '1';

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

            minstep += bfs(i, goal);

            goal++;

        }



        cout << minstep << 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: