// 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: 5.05560149404107x):
#include<iostream>

#include<algorithm>

#include<queue>

const int INF = 100000000;

using namespace std;

typedef pair<int, int> P;

char a[1005][1005];

int N, M;

int sx, sy;

int gx, gy;

int d[1005][1005];

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

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

int bfs()

{

	queue<P> que;

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

		for (int j = 0; j < M; j++)

			d[i][j] = INF;

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

	d[sx][sy] = 0;

	while (que.size())

	{

		P p = que.front();

		que.pop();

		if (p.first == gx&&p.second == gy)

			break;

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

		{

			int nx = p.first + dx[i];

			int ny = p.second + dy[i];

			if (nx >= 0 && nx < N&&ny >= 0 && ny < M&&a[nx][ny] != 'X'&&d[nx][ny] == INF)

			{

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

				d[nx][ny] = d[p.first][p.second] + 1;

			}

		}



	}

	return d[gx][gy];

}

int main(void)

{

	while (cin >> N >> M)

	{

		int n, sum = 0;

		cin >> n;

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

			for (int j = 0; j < M; j++)

				cin >> a[i][j];

		for (int j = 0; j < N; j++)

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

				if (a[j][k] == 'S')

				{

					sx = j;

					sy = k;

				}

		for (int j = 0; j < N; j++)

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

				if (a[j][k] == '1')

				{

					gx = j;

					gy = k;

				}

		sum += bfs();

		for (int i = 0; i < n - 1; i++)

		{

			for (int j = 0; j < N; j++)

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

				{

					if (a[j][k] == (char)(i+49))

					{

						sx = j;

						sy = k;

					}

					if (a[j][k] == (char)(i+50))

					{

						gx = j;

						gy = k;

					}

				}

			sum += bfs();



		}

		cout << sum << endl;

	}

}