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

#include<cstdio>

#include<queue>

#include<utility>

#include<cstring>

using namespace std;

char map[1005][1005];

bool vis[1005][1005];

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

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

int sx,sy;

int h,w;

int bfs(int n){

	memset(vis,false,sizeof(vis));

	queue<pair<pair<int,int>,int> >que;

	que.push(make_pair(make_pair(sx,sy),0));

	while(!que.empty()){

		pair<int,int>now=que.front().first;

		int cnt=que.front().second;

		que.pop();

		if(map[now.first][now.second]==n+48){

			sx=now.first;

			sy=now.second;

			return cnt;

		}

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

			int _x=now.first+dx[i];

			int _y=now.second+dy[i];

			if(_x<0||_y<0||_x>=h||_y>=w||map[_x][_y]=='X'||vis[_x][_y]){

				continue;

			}

			vis[_x][_y]=true;

			que.push(make_pair(make_pair(_x,_y),cnt+1));

		}

	}

}

int main(){

	int n;

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

		memset(map,0,sizeof(map));

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

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

		}

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

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

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

					sx=i;

					sy=j;

					map[i][j]=='.';

					break;

				}

			}

		}

		int count=0;

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

			count+=bfs(k);

		}

		cout<<count<<endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.921620826051455x):
#include<iostream>

#include<cstdio>

#include<vector>

#include<cmath>

#include<map>

#include<set>

#include<queue>

#include<cstring>

#include<algorithm>

using namespace std;

const int maxn = 1000 + 8;

#define inf 0x3f3f3f3f



struct node

{

    int x,y;

    node(int x = 0,int y = 0):x(x),y(y){}

}s,current,st,nd;



int h,w,n;

int step[maxn][maxn];

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

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

string Map[maxn];

int sum = 0;



int bfs()

{

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

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

            step[i][j] = inf;

        }

    }

    queue<node> q;

    step[st.x][st.y] = 0;

    q.push(st);

    while(!q.empty())

    {

        current = q.front();

        q.pop();

        if(current.x == nd.x && current.y == nd.y) break;

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

        {

            s.x = current.x + dx[i];

            s.y = current.y + dy[i];

            if(s.x >= 0 && s.x < h && s.y >= 0 && s.y < w && Map[s.x][s.y] != 'X' && step[s.x][s.y] == inf)

            {

                step[s.x][s.y] = step[current.x][current.y] + 1;

                q.push(s);

            }

        }

    }

    return step[nd.x][nd.y];

}

int main()

{

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

    {

        sum = 0;

        for(int i = 0 ; i < h; ++i) cin >> Map[i];

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

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

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

                    st.x = i;

                    st.y = j;

                }

            }

        }



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

        {

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

            {

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

                {

                    if(Map[j][k] == '1' + i)

                    {

                        nd.x = j; nd.y = k;

                        sum += bfs();

                        st.x = j; st.y = k;

                    }

                }

            }

        }

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

    }

    return 0;

}



/*

3 3 1

S..

...

..1

4 5 2

.X..1

....X

.XX.S

.2.X.

10 10 9

.X...X.S.X

6..5X..X1X

...XXXX..X

X..9X...X.

8.X2X..X3X

...XX.X4..

XX....7X..

X..X..XX..

X...X.XX..

..X.......

*/