// Slower C++ solution:
//??-2015-11-1

# include <cstdio>

# include <string>

# include <cstring>

# include <algorithm>

# include <iostream>

# include <queue>



using namespace std;



int direction[4][2] = { { -1, 0 }, { 1, 0 }, { 0, -1 }, {0,1} };                  //四个方向，上下左右

struct node

{

	int w, h;

	int step;

};

struct node2

{

	int h, w, value;

};

int H, W, N;

string s[1005];

int vis[1005][1005];

int cmp(node2 a, node2 b)

{

	if (a.value < b.value)

		return 1;

	else

		return 0;

}

int DFS(int begin_h, int begin_w, int end_h, int end_w)

{

	int i, j,w = 0;

	queue<node>Q;

	node temp;

	temp.h = begin_h;

	temp.w = begin_w;

	temp.step = 0;

	Q.push(temp);

	vis[begin_h][begin_w] = 1;

	while (!Q.empty())                     //?列不空的?

	{

		node temp1 = Q.front();

		Q.pop();

		//printf("%d\n", w++);

		if (temp1.h == end_h&&temp1.w == end_w)

		{

			return temp1.step;

		}

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

		{

			if (temp1.h + direction[i][0] >= 0 && temp1.h + direction[i][0] < H&&temp1.w + direction[i][1] >= 0 && temp1.w + direction[i][1] < W&&!vis[temp1.h + direction[i][0]][temp1.w + direction[i][1]] && s[temp1.h + direction[i][0]][temp1.w + direction[i][1]] == '.')

			{

				node temp2;

				temp2.h = temp1.h + direction[i][0];

				temp2.w = temp1.w + direction[i][1];

				temp2.step = temp1.step + 1;

				vis[temp1.h + direction[i][0]][temp1.w + direction[i][1]] = 1;

				Q.push(temp2);

			}

		}

	}

}

int main(void)

{

	memset(vis, 0, sizeof(vis));

	int i, j,k = 1;

	int all_step = 0;

	int s_h, s_w,e_h,e_w;

	//freopen("in.txt", "r", stdin);

	scanf("%d%d%d", &H, &W, &N);

	node2 node1[20];

	for (i = 0; i < H; i++)

	{

		cin >> s[i];

		for (j = 0; j < s[i].length(); j++)

		{

			if (s[i][j] >= '0'&&s[i][j] <= '9')

			{

				node1[k].h = i;

				node1[k].w = j;

				node1[k].value = s[i][j] - '0';                //保留?

				k++;

				s[i][j] = '.';                                  //?了BFS的?候?理一致

			}

			else if (s[i][j] == 'S')                            //?始?点

			{

				s_h = i;

				s_w = j;

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

			}

		}

	}

		sort(node1+1, node1 + N+1, cmp);                             //先排好序

		node1[0].h = s_h;                                            //?了使下面循?一?

		node1[0].w = s_w;

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

		{

			e_h = node1[i+1].h;                                      //?点的坐?

			e_w = node1[i+1].w;

			s_h = node1[i].h;

			s_w = node1[i].w;

			//printf("step=%d\n", DFS(s_h, s_w, e_h, e_w));

			all_step += DFS(s_h, s_w, e_h, e_w);                       //原来的?点??起始点

			memset(vis, 0, sizeof(vis));

		}

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

	

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 9.28054394820317x):
#include<iostream>

#include<algorithm>

#include<map>

#include<queue>

using namespace std;

#define MAX 1005



int h, w, n;

char city[MAX][MAX];

struct node{

	int x,y,time;

} start;

int direc1[4] = { -1,0,1,0 }, direc2[4] = { 0,1,0,-1 };

map<int,int>visited;



void dfs(char end) {

	queue<node> que;

	que.push(start);

	visited[start.x*w+start.y]=1;

	int nextx,nexty;

	while (que.size()!=0)

	{

		node now=que.front();

		que.pop();

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

			nextx=now.x + direc1[i];

			nexty=now.y+direc2[i];

			if (nextx >= 0 && nextx< h&&nexty>= 0 && nexty < w && city[nextx][nexty]!='X' && !visited[nextx*w+nexty]) {	



				if (city[nextx][nexty]==end)

				{

					start.x=nextx;

					start.y=nexty;

					start.time=now.time+1;

					return ;

				}

				else{

					visited[nextx*w+nexty]=1;

					node now1;

					now1.x=nextx;

					now1.y=nexty;

					now1.time=now.time+1;

					que.push(now1);

				}

			}

		}

	}



}







int main() {

	cin >> h >> w >> n;	

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

	{

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

		{

			cin >> city[i][j];

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

				start.x= i;

				start.y = j;

				start.time=0;

			}

		}

	}

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

		visited.clear();

		dfs(i+'1');

	}

	cout << start.time << endl;

	//system("pause");

	return 0;

}