// 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: 2.0049277407483257x):
#include <iostream>

#include <queue>

#include <string>

#include <cstdio>

#include <cstring>

#include <utility>

using namespace std;

typedef pair<int, int> P;

char maze[1005][1005];

int d[1005][1005];

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

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

int h, w, n, hard; 

int si, sj;

const int INF = 0x3f3f3f3f;

int tot = 0; 



void BFS(int ai, int aj){

    queue<P> que;

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

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

            d[i][j] = INF;

        }

    }

    que.push(P(ai, aj));

    d[ai][aj] = 0;

    while(!que.empty()){

        P a = que.front();

        que.pop();

        int x = a.first, y = a.second;

        if(maze[x][y]-'0' == hard){

            tot += d[x][y];

            si = x; sj = y;

            break;

        }

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

            int dx = x+dirx[i], dy = y+diry[i];

            if(dx>=0 && dx<h && dy>=0 && dy<w && maze[dx][dy]!='X' && d[dx][dy]==INF){

                d[dx][dy] = d[x][y]+1;

                que.push(P(dx, dy));

            }

        }

    }

}



int main(int argc, char const *argv[]){

    while(scanf("%d%d%d", &h, &w, &n) != EOF){

        getchar();

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

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

                scanf("%c", &maze[i][j]);

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

                    si = i; sj = j;

                }

            }

            getchar();

        }

        tot = 0;

        hard = 1;

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

            BFS(si, sj);

            hard++;

        }

        cout << tot << endl;

    }

    return 0;

}