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

#include<queue>

#include<cctype>





using namespace std;



int cnt;

char field[1002][1002];

int cost[1002][1002];

int factory[9][2];

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

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



int bfs(int sx, int sy, int gx, int gy) {

    queue< pair<int,int> > que;



    for(int i = 0; i < 1002; i++) {

        for (int j = 0; j < 1002; j++) {

            cost[i][j] = -1;

        }

    }



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

    cost[sx][sy] = 0;



    while(!que.empty()) {

        pair<int, int> p;

        p = que.front();

        que.pop();

        

        if(p.first == gx && p.second == gy) {

            break;

        }



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

            int tx = p.first + x[i]; int ty = p.second + y[i];

            if(field[tx][ty] != 'X' && cost[tx][ty] == -1) {

                que.push(pair<int, int>(tx, ty));

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

            }

        }

    }

    return cost[gx][gy];

}



int main() {

    int w, h, n, sx, sy;



    cin >> h >> w >> n;



    for (int y = 1; y <= h; y++) {

        for (int x = 1; x <= w; x++) {

            cin >> field[x][y];

            if (field[x][y] == 'S') {

                sx = x;

                sy = y;

            } else if(isdigit(field[x][y])) {

                int tmp = field[x][y] - '0';

                factory[tmp][0] = x;

                factory[tmp][1] = y;

            }

        }

        field[0][y] = 'X';

        field[w + 1][y] = 'X';

    }



    for (int x = 0; x <= w + 1; x++) {

        field[x][0] = 'X';

        field[x][h + 1] = 'X';

    }



    int cnt = 0;

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

        // 次の目的地へ行くで

        cnt += bfs(sx, sy, factory[i][0], factory[i][1]);

        sx = factory[i][0];

        sy = factory[i][1];

    }



    cout << cnt << endl;

}

// Faster C++ solution for the same problem (measured speedup: 2.506382937354749x):
#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;

}