// Slower C++ solution:
#include"stdio.h"

#include"stdlib.h"

#include"string.h"

#include"algorithm"

#include"queue"

using namespace std;

struct mmm

{

	int x,y;

}vv;

int dir[4][2] = {0,-1,-1,0,0,1,1,0};

char xx[1005][1005];

int bb[11][2];

queue <mmm> qq;

int vis[1005][1005];

int bfs(char t)

{

	int sum = 0;

	int pl = 0;

	while(!qq.empty())

	{

		vv = qq.front();

		int x0 = vv.x;

		int y0 = vv.y;

		//printf("%d %d %d\n",x0,y0,vis[x0][y0]);

		qq.pop();

		if(xx[x0][y0] == t && !pl)

		{

			sum = vis[x0][y0];

			pl = 1;

		}

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

		{

			int xt = x0 + dir[i][0];

			int yt = y0 + dir[i][1];

			if(xx[xt][yt] != '#' && xx[xt][yt] != 'X' && vis[xt][yt] == -1)

			{

				vis[xt][yt] = vis[x0][y0] + 1;

				vv.x = xt;

				vv.y = yt;

				qq.push(vv);

			}

		}

	}

	return sum;

}

int main()

{

	int a,b,c,d,x1,y1,sum;

	while(~scanf("%d %d %d",&a,&b,&c))

	{

	//	if(a == 0 && b == 0)

	//	break;

		memset(xx,'#',sizeof(xx));

		getchar();

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

		{

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

			{

				xx[i][j] = getchar();

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

				{

					x1 = i;

					y1 = j;

				}

				else if(xx[i][j] != 'X')

				{

					bb[xx[i][j] - '0'][0] = i;

					bb[xx[i][j] - '0'][1] = j;

				}

			}

			getchar();

		}

		bb[0][0] = x1;

		bb[0][1] = y1;

		sum = 0;

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

		{

			vv.x = bb[i-1][0];

			vv.y = bb[i-1][1];

			memset(vis,-1,sizeof(vis));

			vis[vv.x][vv.y] = 0;

			qq.push(vv);

			sum += bfs(i + '0');

		}

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

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1943771975274993x):
#include <iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<string>

#include<cstdlib>

#include<stack>

#include<queue>

#include <map>

#include<cmath>

#include<vector>

#define MAX_N 1000000

#define INF 0x3f3f3f3f

#define SIZE 1000

using namespace std;







struct note

{

    int x, y;

}start, ende;

typedef pair<int, int> P;

int H, W, N;

char a[1003][1003];

int sum;

int d[1003][1003];

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

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

void bfs(note h, note m)

{

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

    queue<P> que;

    que.push(P(h.x, h.y));

    d[h.x][h.y] = 0;

    while (!que.empty())

    {

        P p = que.front();

        que.pop();

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

        {

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

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

            if (xx >= 0 && xx < H && yy >= 0 && yy < W && a[xx][yy] != 'X' && d[xx][yy] == -1)

            {

                d[xx][yy] = d[p.first][p.second] + 1;

                que.push(P(xx, yy));

            }

        }

    }

    sum += d[m.x][m.y];

    start = ende;

}

void solve()

{

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

    {

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

        {

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

            {

                start.x = i;

                start.y = j;

                i = H;

                j = W;

            }

        }

    }

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

    {

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

        {

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

            {

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

                {

                    ende.x = i;

                    ende.y = j;

                    bfs(start, ende);

                    i = H;

                    j = W;

                }

            }

        }

    }

    cout << sum << endl;

}

int main()

{

    cin >> H >> W >> N;

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

    {

        cin >> a[i];

    }

    sum = 0;

    solve();

    return 0;

}