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

#include <queue>

#include <string.h>

using namespace std;

typedef pair<int, int> P;

int sx,sy,H,W,N;

char map1[1002][1002];

int step[1002][1002];

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

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

void find1(int g){

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

    for(int k = 0; k < H; k++)

        if(map1[i][k] == 'S' || map1[i][k] == (g+48)){

        if(map1[i][k] == 'S') map1[i][k] ='.';

        sx = i; sy = k;

        return ;

    }

}



int bfs(int n)

{

    memset(step,-1,sizeof(step));

    queue <P> que;

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

    step[sx][sy] = 0;

    while(que.size()){

        int x = que.front().first, y = que.front().second;

        que.pop();

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

            int xn = x + dx[i], yn = y + dy[i];

            if(xn >= 0 && xn < W && yn >= 0 && yn < H && map1[xn][yn] != 'X' && step[xn][yn] == -1)

                {

                    step[xn][yn] = step[x][y] + 1;

                    if(map1[xn][yn] == (n+48))

                        {

                            return step[xn][yn];

                        }

                    que.push(P(xn, yn));

                }

        }

    }

return 0;

}



int main()

{

    cin >> W >> H >> N;

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

        for(int k = 0; k < H; k++)

        cin >> map1[i][k];

    int step = 0;

    find1(0);

    step += bfs(1);

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

    {



        find1(i);

        step += bfs(i+1);

    }

    cout << step << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.050422444824758x):
#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;

}