// 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: 1.7681737537873794x):
#include <iostream>

#include <queue>

#include <string>

using namespace std;

int main()

{

	int h,w,n;

	int sx,sy;

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

	string map[1001];

	int d[1001][1001];

	int x = 0;

	cin >> h >> w >> n;

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

	{

		cin >> map[i];

	}

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

	{

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

		{

			d[j][i] = -1;

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

			{

				sx = j;

				sy = i;

			}

		}

	}

	d[sx][sy] = 0;

	for(int i=0;i<9;i++)

	{

		queue< pair<int, int> > que;

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

		{

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

			{

				d[k][j] = -1;

			}

		}

		d[sx][sy] = 0;

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

		while(que.size())

		{

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

			que.pop();

			if((map[p.second][p.first] - '0') == (i+1))

			{

				sx = p.first;

				sy = p.second;

				x+=d[sx][sy];

				break;

			}

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

			{

				int nx = p.first + dx[j];

				int ny = p.second + dy[j];

				if(0 <= nx && nx < w && 0 <= ny && ny < h && map[ny][nx] != 'X' && d[nx][ny] == -1)

				{

					que.push(make_pair(nx,ny));

					d[nx][ny] = d[p.first][p.second] + 1;

				}

			}

		}

	}

	cout << x << endl;

}