// Original C++ solution (slower):
#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

const int maxn=1005;

int n,m,k,dx[]={-1,1,0,0},dy[]={0,0,1,-1},vis[maxn][maxn];

char map[maxn][maxn];

struct maze

{

    maze(int xx=0,int yy=0,int tt=0):x(xx),y(yy),time(tt){}

	int x,y,time;

};

maze bfs(int bx,int by,char ch)

{

	memset(vis,0,sizeof(vis));

	queue<maze> q;

	maze st(bx,by,0);

	q.push(st);vis[bx][by]=1;

	while(!q.empty())

	{

		st=q.front();q.pop();

		if(map[st.x][st.y]==ch)

		{

			return st;

		}

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

		{

			int tx=st.x+dx[i],ty=st.y+dy[i];

			if(tx<0||tx>=n||ty<0||ty>=m||map[tx][ty]=='X')//越界和?

			{

				continue;

			}

			maze tp(tx,ty,st.time+1);

			if(!vis[tx][ty])

			{

				vis[tx][ty]=1;

				q.push(tp);

			}

		}

	}

}

int main()

{



	while(~scanf("%d%d%d",&n,&m,&k))

	{

		int bx,by;

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

		{

			scanf("%s",map[i]);

			for(int j=0;j<m;++j)

			{

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

				{

					bx=i;by=j;

					

					map[i][j]='0'; //第0个起点

				}

			}

		}

		int ans=0;

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

		{

			maze tp=bfs(bx,by,map[bx][by]+1);

			ans+=tp.time;

			bx=tp.x;by=tp.y;

		}

		printf("%d\n",ans);

	}

	return 0;

}

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

#include <cstring>

#include <queue>

#include <cctype>



using namespace std;



struct State {

    int step;

    int x;

    int y;



    State() {

        step = 0;

        x = 0;

        y = 0;

    }



    State(int step, int x, int y): step(step), x(x), y(y) {}

};



const int MAX_N = 1000;

const int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, { -1, 0}};

char board[MAX_N][MAX_N];

int W, H, N;

State pos[10];



void init();

void input();

void solve();

int bfs(State from, State to);

void findPossition();



int main() {

#ifdef LOCAL

    freopen("data.in", "r", stdin);

    freopen("data.out", "w", stdout);

#endif

    init();

    input();

    solve();

    return 0;

}



void init() {

}



void input() {

    scanf("%d %d %d", &H, &W, &N);



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

        scanf("%s", board[i]);

    }

}





void solve() {

    int moves = 0;

    findPossition();



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

        moves += bfs(pos[i], pos[i + 1]);

    }



    printf("%d\n", moves);

}



int bfs(State from, State to) {

    queue<State> que;

    bool visit[H][W];

    State cur = from;



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

        memset(visit[i], false, sizeof(visit[i]));

    }



    que.push(cur);



    while(!que.empty()) {

        cur = que.front();

        que.pop();



        if(visit[cur.x][cur.y]) {

            continue;

        }

        

        visit[cur.x][cur.y] = true;



        if(cur.x == to.x && cur.y == to.y) {

            break;

        }



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

            int x = cur.x + dir[i][0];

            int y = cur.y + dir[i][1];



            if(x < 0 || x >= H || y < 0 || y >= W || visit[x][y] || 'X' == board[x][y]) {

                continue;

            }



            que.push(State(cur.step + 1, x, y));

        }

    }



    return cur.step;

}



void findPossition() {

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

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

            int c = board[i][j];



            if('S' == c) {

                pos[0] = State(0, i, j);

            }



            if(isdigit(c)) {

                pos[c - '0'] = State(0, i, j);

            }

        }

    }

}

// Fastest known solution for this problem: