// 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: 2.524149428908771x):
#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <vector>

#include <queue>



using namespace std;

struct as

{

    int x,y;

}pos[10]; //用来存储每次的起点和终点

struct node

{

    int x,y;

    int step;

};

bool vis[1010][1010];

char maze[1050][1050];

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

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

int H,W,N;



bool check(int x,int y)

{

    if (x<=0 || x>H) return 1;

    if (y<=0 || y>W) return 1;

    if (vis[x][y]) return 1;

    if (maze[x][y] =='X')return 1;

    return 0;

}

int bfs(int m,int n)

{

    memset(vis,0,sizeof vis);

    int sx = pos[m].x, sy = pos[m].y;

    int gx = pos[n].x, gy = pos[n].y;

   // printf("%d %d\n",sx,sy);

   // printf("%d %d\n",gx,gy);

    queue<node> Q;

    node a,next;

    a.step = 0;

    a.x = sx, a.y =sy;

    Q.push(a);

    while(!Q.empty())

    {

        a = Q.front();

        Q.pop();

        if (a.x == gx && a.y == gy)

            return a.step;

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

        {

            int nx = a.x + dx[i];

            int ny = a.y + dy[i];

            if(check(nx,ny))

                continue;

            else

            {

                vis[nx][ny] =1;

                next.x = nx,next.y = ny,next.step = a.step+1;

                Q.push(next);

            }

        }

    }

    return 999999;

}



int main()

{

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

    while(cin>>H>>W>>N)

    {

        int sx,sy;

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

        {

            for(int j = 1; j <= W;j++)

            {

                cin>>maze[i][j];

                if (maze[i][j] >='1' && maze[i][j]<='9')

                {

                    pos[maze[i][j]-'0'].x = i;

                    pos[maze[i][j]-'0'].y = j;

                }

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

                    sx=i,sy = j;

            }

        }



        pos[0].x = sx,pos[0].y = sy;

        int ans = 0;

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

            ans += bfs(i,i+1);

        cout<<ans<<endl;

    }

    return 0;

}