// Slower C++ solution:
#include <iostream>

#include <cmath>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>



#define INF 0x7f7f7f7f

#define PI acos(-1.0)



#define L(x)    (x) << 1

#define R(x)    (x) << 1 | 1

#define lowbit(x)   (x)&(-x)

#define Mid(l, r)   (l + r) >> 1

#define Min(x, y)   (x) < (y) ? (x) : (y)

#define Max(x, y)   (x) < (y) ? (y) : (x)

#define iabs(x)     (x) < 0 ? -(x) : (x)

#define CLR(arr, val)    memset(arr, val, sizeof(arr))



using namespace std;



char map[1024][1024];

bool vis[1024][1024];

int cur, h, w;

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



struct node {

	int x, y, step;

} sta, now;



bool check(int x, int y) {

	if (x >= 0 && x < h && y >= 0 && y < w && !vis[x][y] && map[x][y] != 'X')	return true;

	return false;

}







void bfs(node s) {

	memset(vis, 0, sizeof(vis));



	queue<node> que;

	s.step = 0;

	que.push(s);

	node next;

	while (!que.empty()) {

//		cout << "h\n";

		s = que.front();

//		cout << s.x << ' ' << s.y << ' ' << cur << endl;

		que.pop();

		if (map[s.x][s.y] == cur + '0')	{

			sta = s;

			return ;

//			s.step;

		}



//		cout << "a\n";

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

//			cout << i << endl;

			next.x = s.x + dir[i][0];

			next.y = s.y + dir[i][1];

			next.step = s.step + 1;

			if (check(next.x, next.y)) {

				vis[next.x][next.y] = true;

				que.push(next);

			}



		}

	}

}



int main() {

	int n;

	while (cin >> h >> w >> n) {

//		memset(map, 0, sizeof(map));

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

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

				cin >> map[i][j];

				if (map[i][j] == 'S')	sta.x = i, sta.y = j, map[i][j] = '.';

			}

		}

		sta.step = 0;

		int ans = 0;

		for (cur = 1; cur <= n; cur++) {

			bfs(sta);

			ans += sta.step;

			//cout << ans << endl;

		}

		cout << ans << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.043633088949998x):
#include<iostream>

#include<iomanip>

#include<cstdlib>

#include<math.h>

#include<algorithm>

#include<string>

#include<vector>

#include<list>

#include<queue>

#include<stack>

#include<deque>

#include<set>

#include<map>

#define REP(i, N) for(ll i = 0; i < N; ++i)

#define FOR(i, a, b) for(ll i = a; i < b; ++i)

#define ALL(a) (a).begin(),(a).end()

#define pb push_back

#define INF 1000000000

#define MOD 1000000007

using namespace std;

typedef long long ll;

typedef pair<ll, ll> P;

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

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

int qx[8] = {1, 1, 0, -1, -1, -1, 0, 1};

int qy[8] = {0, 1, 1, 1, 0, -1, -1, -1};



int main(void) {

	ll h, w, n;

	cin>>h>>w>>n;

	vector<string> a(h);

	vector<vector<ll> > dist(h, vector<ll>(w, INF));

	ll time = 0;

	P sp;

	REP(i, h) cin>>a[i];

	bool det = false;

	REP(i, h) {

		REP(j, w) {

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

				sp.first = i;

				sp.second = j;

				det = true;

				break;

			}

		}

		if(det) break;

	}

	REP(i, n) {

		queue<P> c;

		c.push(sp);

		dist[sp.first][sp.second] = 0;

		while(true) {

			P cp = c.front();

			c.pop();

			if(a[cp.first][cp.second] - '1' == i) {

				time += dist[cp.first][cp.second];

				sp = cp;

				break;

			}

			REP(j, 4) {

				ll nx = cp.first + dx[j];

				ll ny = cp.second + dy[j];

				if(nx >= 0 && nx < h && ny >= 0 && ny < w && dist[nx][ny] == INF && (a[nx][ny] == 'S' || a[nx][ny] == '.' || (a[nx][ny] >= '1' && a[nx][ny] <= '9'))) {

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

					dist[nx][ny] = dist[cp.first][cp.second] + 1;

				}

			}

		}

		REP(j, h) {

			REP(k, w) {

				dist[j][k] = INF;

			}

		}

	}

	cout<<time<<endl;

}