// Slower C++ solution:
/************************************************

* Author        :somniloquy

* Created Time  :2015/10/30 16:15:09

 ************************************************/



#include <cstdio>

#include <algorithm>

#include <iostream>

#include <cstring>

#include <queue>



using namespace std;



const int hw_max = 1000 + 5;

char graph[hw_max][hw_max];

int sign[hw_max][hw_max];

int h, w, n, ans;

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

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

struct node

{

	int x;

	int y;

	int time;

} start;



void BFS(node op, char ed)

{

	queue <node> process;

	process.push(op);

	sign[op.x][op.y] = 1;

	while(!process.empty())

	{

		node now = process.front();

		process.pop();

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

		{

			node next = now;

			next.x += dx[i];

			next.y += dy[i];

			next.time ++;

			if(next.x >= 0 && next.x < h && next.y >= 0 && next.y < w && graph[next.x][next.y] != 'X' && !sign[next.x][next.y])

			{

				if(graph[next.x][next.y] == ed)

				{

					ans += next.time;

					next.time = 0;

					start = next;

				       	return;

				}

				else

				{

					process.push(next);

					sign[next.x][next.y] = 1;

				}

			}

		}

	}

}



int main()

{

	while(~scanf("%d %d %d", & h, & w, & n))

	{

		getchar();

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

		{

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

			{

			       scanf("%c", & graph[i][j]);

			       if(graph[i][j] == 'S')

			       {

					start.x = i;

					start.y = j;

					start.time = 0;

			       }

			}

			getchar();

		}

		ans = 0;

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

		{

			memset(sign, 0, sizeof(sign));

			BFS(start, '0' + i);

		}

		printf("%d\n", ans);

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.07698095516622x):
#include <iostream>

#include <queue>

using namespace std;

int w, h, n;

char map[1024][1024];

// 各点到当前工厂的距?

int d[1024][1024];

const int direction[4][2] = {

	{ -1, 0 },

	{ 1, 0 },

	{ 0, -1 },

	{ 0, 1 },

};

 

int factory[16][2];

typedef pair<int, int> P;



int bfs(const int& sx, const int& sy, const int& gx, const int& gy)

{

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

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

	{

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

		{

			d[j][i] = -1;

		}

	}

	queue<P> que;

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

	d[sx][sy] = 0;

	while (que.size())

	{

		const P p = que.front(); que.pop();

		// 如果是?点就?束

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

		{

			break;

		}

 

		// 四方向漫游

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

		{

			int nx = p.first + direction[i][0];

			int ny = p.second + direction[i][1];

			// 是否可以移?，并且?点没有???

			if (0 <= nx && nx < w && 0 <= ny && ny < h && map[nx][ny] != 'X' && d[nx][ny] == -1)

			{

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

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

			}

		}

	}



	return d[gx][gy];

}

 

///////////////////////////SubMain//////////////////////////////////

int main(int argc, char *argv[])

{

 

	cin >> h >> w >> n;

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

	{

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

		{

			cin >> map[j][i];

		}

	}

 

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

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

		{

			if (map[j][i] == 'S')

		{

				factory[0][0] = j;

				factory[0][1] = i;

				map[j][i] = '.';

			}

			else if (isdigit(map[j][i]))

			{

				int index = map[j][i] - '0';

				factory[index][0] = j;

				factory[index][1] = i;

			}

		}

	}

 

	int step = 0;

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

	{

		// 按?序吃遍中?

		step += bfs(factory[i][0], factory[i][1], factory[i + 1][0], factory[i + 1][1]);

	}

	

	cout << step << endl;

 

	return 0;

}