// 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: 2.0035472980437388x):
#include<iostream>

#include<algorithm>

#include<queue>

const int INF = 100000000;

using namespace std;

typedef pair<int, int> P;

char a[1005][1005];

int N, M;

int sx, sy;

int gx, gy;

int d[1005][1005];

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

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

int bfs()

{

	queue<P> que;

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

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

			d[i][j] = INF;

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

	d[sx][sy] = 0;

	while (que.size())

	{

		P p = que.front();

		que.pop();

		if (p.first == gx&&p.second == gy)

			break;

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

		{

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

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

			if (nx >= 0 && nx < N&&ny >= 0 && ny < M&&a[nx][ny] != 'X'&&d[nx][ny] == INF)

			{

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

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

			}

		}



	}

	return d[gx][gy];

}

int main(void)

{

	while (cin >> N >> M)

	{

		int n, sum = 0;

		cin >> n;

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

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

				cin >> a[i][j];

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

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

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

				{

					sx = j;

					sy = k;

				}

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

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

				if (a[j][k] == '1')

				{

					gx = j;

					gy = k;

				}

		sum += bfs();

		for (int i = 0; i < n - 1; i++)

		{

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

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

				{

					if (a[j][k] == (char)(i+49))

					{

						sx = j;

						sy = k;

					}

					if (a[j][k] == (char)(i+50))

					{

						gx = j;

						gy = k;

					}

				}

			sum += bfs();



		}

		cout << sum << endl;

	}

}