// 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: 4.692606401677613x):
#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <queue>

using namespace std;



const int N = 1000;

const int INF = 0x3f3f3f3f;



int n, m, f;

char map[N][N+1];



bool legal(int x, int y, int k)

{

    return x >= 0 && x < n && y >= 0 && y < m

        && map[x][y] != 'X';

}



void input()

{

    cin >> n >> m >> f;

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

        scanf("%s", map[i]);

    }

}



int BFS(int k)

{

    int i, j;

    int d[N][N];

    typedef pair<int, int> P;

    queue <P> q;

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

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

            d[i][j] = INF;

            if ((k == 0 && map[i][j] == 'S') || map[i][j] == '0'+k) {

                q.push(P(i, j));

                d[i][j] = 0;

                map[i][j] = '.';

            }

        }

    }



    while( q.size() ) {

        P p = q.front();

        q.pop();

        int x = p.first;

        int y = p.second;



        if (map[x][y] == INF)

            return d[x][y];



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

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

            int nx = x + pos[i][0];

            int ny = y + pos[i][1];

            if (legal(nx, ny, k) && d[nx][ny] == INF) {

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

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

                if (map[nx][ny] == '0'+k+1) {

                    return d[nx][ny];

                }

            }

        }

    }



    return INF;

}



void solve()

{

    int res = 0;

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

        res += BFS(i);

    }

    printf("%d\n", res);

}



int main(void)

{

    input();



    solve();



    return 0;

}