// 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: 5.095014382969503x):
#include <iostream>

#include <stdio.h>

#include <deque>



using namespace std;



const int INF = 0x3f3f3f3f;

typedef pair<int, int> P;

int h, w, n;

int sx, sy;

int gx, gy;

char cmap[1000][1000];

int goalSum;

int goal[10];

int d[1000][1000];

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

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



int bfs() {

    int res = 0;

    deque<P> que;

    for (int ci = 0; ci < n; ci++) {  //要?行n次循?，?找n次?点，?次?数?入goal[i], 最后累加起来是res 

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

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

                d[i][j] = INF;

            }

        }

        

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

        d[sx][sy] = 0;

        

        while(que.size()) {            

            int x = que.front().first;

            int y = que.front().second;

                        

            if ((int)cmap[x][y] == 49 + ci) {

                goal[ci] = d[x][y];

                sx = x;

                sy = y;

                break;

            }

            

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

                int nx = x + dx[i];

                int ny = y + dy[i];

                if (nx >=0 && nx < h && ny >= 0 && ny < w && cmap[nx][ny] != 'X' && d[nx][ny] == INF) {

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

                    d[nx][ny] = d[x][y] + 1;

                }

            }    

            que.pop_front();        

        }

        que.clear(); 

        res += goal[ci];

    }

    return res;

}



int main() {

    cin >> h >> w >> n;

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

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

            cin >> cmap[i][j];

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

                sx = i;

                sy = j;

            }

        }

    }

    goalSum = bfs();

    cout << goalSum << endl;

    return 0;

}