// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

const int MAX = 1024;



int h, w, n, maze[MAX][MAX];

int b[15];

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

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

int d[MAX][MAX];

int ans = 0;



int readchar() {

    for(;;) {

        int ch = getchar();

        if(ch != '\n' && ch != '\r') return ch;

    }

}



int bfs(int x, int y, int x2, int y2)

{

    //cout << x << y << x2 << y2 << endl;



    queue<int> q;

    q.push(x * 10000 + y);

    while(!q.empty()) {

        int v = q.front();

        q.pop();

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

            int x1 = v / 10000 + r[i];

            int y1 = v % 10000 + z[i];

            if(x1 == x2 && y1 == y2) {

                d[x1][y1] = d[v / 10000][v % 10000] + 1;

                return 0;

            }

            if(x1 >= 0 && y1 >= 0 && x1 < h && y1 < w && maze[x1][y1] != 'X' && d[x1][y1] == -1) {

                 d[x1][y1] = d[v / 10000][v % 10000] + 1;

                 q.push(x1 * 10000 + y1);

            }

        }

    }

    return 0;

}

int main()

{

    cin >> h >> w >> n;

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

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

             maze[i][j] = readchar();

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

            b[0] = i * 10000 + j;

        }

        if(maze[i][j] != 'S' && maze[i][j] != 'X' && maze[i][j] != '.') {

            b[maze[i][j] - '0'] = i * 10000 + j;

        }

    }

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

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

        d[b[i] / 10000][b[i] % 10000] = 0;

        bfs(b[i] / 10000, b[i] % 10000, b[i + 1] / 10000, b[i + 1] % 10000);

        ans += d[b[i + 1] / 10000][ b[i + 1] % 10000];

    }

    cout << ans << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.239235402871074x):
#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;

}