// Slower C++ solution:
//

//  AOJ 0558 Chees

//

//  Created by TaoSama on 2015-02-20

//  Copyright (c) 2014 TaoSama. All rights reserved.

//

#include <algorithm>

#include <cctype>

#include <cmath>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iomanip>

#include <iostream>

#include <map>

#include <queue>

#include <string>

#include <set>

#include <vector>

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



using namespace std;

const int INF = 0x3f3f3f3f;

const int MOD = 1e9 + 7;

const int N = 1e5 + 10;



struct Point {

	int x, y;

	Point(int x = 0, int y = 0): x(x), y(y) {}

} c[15];

int n, m, k, dp[1005][1005];

int d[4][2] = { -1, 0, 1, 0, 0, 1, 0, -1};

char a[1005][1005];



int bfs(int x) {

	memset(dp, 0x3f, sizeof dp);

	queue <Point> q; q.push(c[x - 1]);

	dp[c[x - 1].x][c[x - 1].y] = 0;

	while(!q.empty()) {

		Point cur = q.front(); q.pop();

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

			Point nxt(cur.x + d[i][0], cur.y + d[i][1]);

			if(nxt.x < 1 || nxt.x > n || nxt.y < 1 || nxt.y > m

			        || a[nxt.x][nxt.y] == 'X') continue;

			if(dp[nxt.x][nxt.y] == INF) {

				dp[nxt.x][nxt.y] = dp[cur.x][cur.y] + 1;

				q.push(nxt);

			}

		}

	}

	return dp[c[x].x][c[x].y];

}

int main() {

#ifdef LOCAL

	freopen("in.txt", "r", stdin);

//	freopen("out.txt","w",stdout);

#endif

	ios_base::sync_with_stdio(0);



	while(cin >> n >> m >> k) {

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

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

				cin >> a[i][j];

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

					c[0].x = i, c[0].y = j;

				if(isdigit(a[i][j]))

					c[a[i][j] - '0'].x = i, c[a[i][j] - '0'].y = j;

			}

		}

		int ans = 0;

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

			ans += bfs(i);

		cout << ans << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.091262952126499x):
#include <iostream>

#include <cstdio>

#include <queue>

#include <algorithm>

#include <cstring>

#include <string>

#include <stack>

#include <stdlib.h>

#include <cctype>

#include <vector>

#include <map>

#include <set>

#include <cmath>

#include <iomanip>

using namespace std;

const int INF=1000000000;

const int max_n=1005;

typedef pair<int,int>P;

char maze[max_n][max_n+1];

int n,m,sx,sy;

int d[max_n][max_n];

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

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

int bfs(char e)

{

    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.empty())

    {

        P q=que.front();

        que.pop();

        if(maze[q.first][q.second]==e){sx=q.first;sy=q.second;break;}

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

        {

            int nx=q.first+dx[i],ny=q.second+dy[i];

            if(0<=nx&&nx<n&&ny>=0&&ny<m&&maze[nx][ny]!='X'&&d[nx][ny]==INF)

            {

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

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

            }

        }

    }

return d[sx][sy];

}

int main()

{

    int t;

    while(cin>>n>>m>>t)

    {

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

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

       {

           cin>>maze[i][j];

           if(maze[i][j]=='S')sx=i,sy=j;

       }



       int res=0;

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

        res+=bfs(i+'0');

       }

        cout<<res<<endl;

    }

    return 0;

}