// 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: 4.615349720722101x):
#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <vector>

#include <queue>



using namespace std;

struct as

{

    int x,y;

}pos[10]; //用来存??次的起点和?点

struct node

{

    int x,y;

    int step;

};

bool vis[1010][1010];

char maze[1050][1050];

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

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

int H,W,N;



bool check(int x,int y)

{

    if (x<=0 || x>H) return 1;

    if (y<=0 || y>W) return 1;

    if (vis[x][y]) return 1;

    if (maze[x][y] =='X')return 1;

    return 0;

}

int bfs(int m,int n)

{

    memset(vis,0,sizeof vis);

    int sx = pos[m].x, sy = pos[m].y;

    int gx = pos[n].x, gy = pos[n].y;

   // printf("%d %d\n",sx,sy);

   // printf("%d %d\n",gx,gy);

    queue<node> Q;

    node a,next;

    a.step = 0;

    a.x = sx, a.y =sy;

    Q.push(a);

    while(!Q.empty())

    {

        a = Q.front();

        Q.pop();

        if (a.x == gx && a.y == gy)

            return a.step;

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

        {

            int nx = a.x + dx[i];

            int ny = a.y + dy[i];

            if(check(nx,ny))

                continue;

            else

            {

                vis[nx][ny] =1;

                next.x = nx,next.y = ny,next.step = a.step+1;

                Q.push(next);

            }

        }

    }

    return 999999;

}



int main()

{

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

    while(cin>>H>>W>>N)

    {

        int sx,sy;

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

        {

            for(int j = 1; j <= W;j++)

            {

                cin>>maze[i][j];

                if (maze[i][j] >='1' && maze[i][j]<='9')

                {

                    pos[maze[i][j]-'0'].x = i;

                    pos[maze[i][j]-'0'].y = j;

                }

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

                    sx=i,sy = j;

            }

        }



        pos[0].x = sx,pos[0].y = sy;

        int ans = 0;

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

            ans += bfs(i,i+1);

        cout<<ans<<endl;

    }

    return 0;

}



/*

入力例 2

4 5 2

.X..1

....X

.XX.S

.2.X.



出力例 2

12

*/