// 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: 5.084489751548139x):
#include <iostream>

#include <queue>

#include <string>



using namespace std;



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

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

int First = 1000;

int w,h,n;

int ans = 0;

int d[1001][1001];

typedef struct p{

	int x, y;

}P;

string mapp[1001];

P nezumi[11];





//幅優先探索を行う

void bfs(){

	for(int i=1 ; i <= n ; i++ ){

		P sp = nezumi[i-1];

		int sx = nezumi[i-1].x;

		int sy = nezumi[i-1].y;

		int gx = nezumi[i].x;

		int gy = nezumi[i].y;



		//二次元配列の初期化をする

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

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

				d[y][x] = First;

			}

		}

		

		queue<P> q;

		q.push( sp );

		d[sy][sx] = 0;

		while( !q.empty() ){

			P now = q.front();

			q.pop();

			int x = now.x;

			int y = now.y;



			if( x == gx && y == gy ){

				ans += d[gy][gx];



				/*動作確認

				cout << "\n";

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

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

						if (d[y][x]==First){cout << "X";}

							else{cout << d[y][x];}

					}

					cout << "\n";

				}*/

				break;

			}



			//redandblackと同じように探索

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

				int newx = x + dx[i];

				int newy = y + dy[i];

				if( newx < 0 || newy < 0 || newx >= w || newy >= h ) continue;

				if( mapp[newy][newx] != 'X' ){

					P next;

					next.x = newx;

					next.y = newy;

					if( d[newy][newx] == First ){

						d[newy][newx] = d[y][x] + 1;

						q.push( next );

					}

				}

			}

		}

	}

}



int main(){

	//標準入力

	cin >> h >> w >> n;

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

		cin >> mapp[y];

	}

	

	//スタートと数字の確認

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

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

			char c = mapp[y][x];



			//アスキー使っていい感じにしてみる

			if( c >= '1' && c <= '9' ){

				nezumi[ c - '0' ].x = x;

				nezumi[ c - '0' ].y = y;

			}

			if(c == 'S'){

				nezumi[0].x = x;

				nezumi[0].y = y;

			}

		}

	}

	bfs();

	cout << ans << endl;

}