// 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: 1.9465958912361738x):
#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.......

*/