// Slower C++ solution:
#include <bits/stdc++.h>

/*

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/rb_tree>

*/

#define mem(x,y) memset(x,y,sizeof(x))

//#define P(x) cout<<#x<<" = "<<x<<endl;

#define PN(x,n) cout<<'{'<<x[0];for(int i=1;i<n;i++)cout<<','<<x[i];cout<<'}'<<endl;

#define RP(i,n) for(int i=0;i<n;i++)

#define RP2(i,a,b) for(int i=a;i<=b;i++)

#define mp make_pair

#define P pair<int,int>

#define pb push_back

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define dinf 1e99

#define OJ

#define ite iterator

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

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

using namespace std;/*

using namespace __gnu_cxx;

using namespace __gnu_pbds;

typedef long long ll;

typedef unsigned long long ull;

typedef tree<int, int, less<int>, rb_tree_tag, tree_order_statistics_node_update> Tree;

const double PI = acos(-1.0);  

const double eps = 1e-5;

const double T = 5; //t of fire

const double delta = 0.99; //decreasing speed of t

const int INF = 1000000000;*/

const int maxn=1005;

bool vis[maxn][maxn];

int d[maxn][maxn];

int ans,n,m,k,sx,sy;

int lx[maxn],ly[maxn],G[maxn][maxn];

char bf;

queue<P> q;

int bfs(int sx,int sy,int ex,int ey){

    mem(vis,0);

    q.push(mp(sx,sy));

    vis[sx][sy] = 1;

    d[sx][sy] = 0;

    while(!q.empty()){

	P u = q.front(); q.pop();

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

	    P v=mp(u.fi+dx[dr],u.se+dy[dr]);

	    if(v.fi<1 || v.se<1 || v.fi>n || v.se>m || vis[v.fi][v.se] || G[v.fi][v.se]) continue;

	    d[v.fi][v.se]=d[u.fi][u.se] + 1;

	    vis[v.fi][v.se] = 1;

	    q.push(v);

	}

    }

    return d[ex][ey];

}

int main(){

    cin>>n>>m>>k;

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

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

	    cin>>bf;

	    if(bf=='.') G[i][j] = 0;

	    else if(bf == 'S') sx = i, sy = j, G[i][j] = 0;

	    else if(bf=='X') G[i][j] = 1;

	    else lx[bf-48] = i, ly[bf-48]=j;

	}

    }

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

	ans += bfs(sx,sy,lx[i],ly[i]);

	sx=lx[i], sy=ly[i];

    }

    cout<<ans<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.9423788317900144x):
#include <iostream>

#include <queue>

#include <string>



using namespace std;



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

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

int First = 1000;

int w,h,n;

int ans = 0;

int d[1001][1001];

typedef struct p{

	int x, y;

}P;

string mapp[1001];

P nezumi[11];





//幅優先探索を行う

void bfs(){

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

		P sp = nezumi[i-1];

		int sx = nezumi[i-1].x;

		int sy = nezumi[i-1].y;

		int gx = nezumi[i].x;

		int gy = nezumi[i].y;



		//二次元配列の初期化をする

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

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

				d[y][x] = First;

			}

		}

		

		queue<P> q;

		q.push( sp );

		d[sy][sx] = 0;

		while( !q.empty() ){

			P now = q.front();

			q.pop();

			int x = now.x;

			int y = now.y;



			if( x == gx && y == gy ){

				ans += d[gy][gx];



				/*動作確認

				cout << "\n";

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

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

						if (d[y][x]==First){cout << "X";}

							else{cout << d[y][x];}

					}

					cout << "\n";

				}*/

				break;

			}



			//redandblackと同じように探索

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

				int newx = x + dx[i];

				int newy = y + dy[i];

				if( newx < 0 || newy < 0 || newx >= w || newy >= h ) continue;

				if( mapp[newy][newx] != 'X' ){

					P next;

					next.x = newx;

					next.y = newy;

					if( d[newy][newx] == First ){

						d[newy][newx] = d[y][x] + 1;

						q.push( next );

					}

				}

			}

		}

	}

}



int main(){

	//標準入力

	cin >> h >> w >> n;

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

		cin >> mapp[y];

	}

	

	//スタートと数字の確認

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

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

			char c = mapp[y][x];



			//アスキー使っていい感じにしてみる

			if( c >= '1' && c <= '9' ){

				nezumi[ c - '0' ].x = x;

				nezumi[ c - '0' ].y = y;

			}

			if(c == 'S'){

				nezumi[0].x = x;

				nezumi[0].y = y;

			}

		}

	}

	bfs();

	cout << ans << endl;

}