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

#include<cstring>

#include<queue>

using namespace std;

int n,m,k,step[1010][1010],temp[1010][3];

char a[1010][1010];

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

bool vis[1010][1010];

typedef pair<int,int> P;

int bfs(int x,int y,int ex,int ey)

{

	memset(vis,true,sizeof(vis));

	memset(step,0,sizeof(step));

	queue <P> que;

	que.push(P(x,y));

	vis[x][y]=false;

	while(!que.empty())

	{

		P tmp=que.front();que.pop();

		int b=que.size();

		if(tmp.first==ex&&tmp.second==ey)

		  return step[tmp.first][tmp.second];

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

		{

			int tx=tmp.first+dx[i];

			int ty=tmp.second+dy[i];

			if(tx>0&&tx<=n&&ty>0&&ty<=m&&vis[tx][ty]==true&&a[tx][ty]!='X')

			{

				que.push(P(tx,ty));

				vis[tx][ty]=false;

				step[tx][ty]=step[tmp.first][tmp.second]+1;

			}

		}

	}

}

int main()

{

	ios::sync_with_stdio(false);

	cin>>n>>m>>k;

		int sx,sy;

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

		  for(int j=1;j<=m;j++)

		  {

			cin>>a[i][j];

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

			  sx=i,sy=j;

			//temp[0][0]=i;

			//temp[0][1]=j;

		  }

		a[sx][sy]='0';

		int tot=0,ans=0;

		for(int i=1;i<=k;i++)

		{

			for(int j=1;j<=n;j++)

			  for(int l=1;l<=m;l++)

			    if(a[j][l]=='0'+i)

			    {

			    	tot=bfs(sx,sy,j,l);

			    	sx=j,sy=l;

				}

			ans+=tot;

		}

		cout<<ans<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.082153284426668x):
#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;

}