// Slower C++ solution:
//

// Create by 神舟 on 2015-02-21

//



#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <cstdlib>

#include <algorithm>

#include <cctype>

#include <stack>

#include <queue>

#include <map>

#include <string>

#include <set>

#include <vector>

using namespace std;



#define CLR(x) memset(x,0,sizeof x)

#define ll long long

#define inf 0x3f3f3f3f

const int maxn=1e3+5;

const int MOD=5e5+5;

int n,m,t,sx,sy,ex,ey;

char word[maxn][maxn];

int path[maxn][maxn];

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



struct pnt

{

    int x,y;

    pnt(int x=0,int y=0):x(x),y(y){};

};



void bfs()

{

    memset(path,0x3f,sizeof path);

    queue <pnt> que;

    pnt p,np;

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

    path[sx][sy]=0;

    que.push(p);

    while(!que.empty()){

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

        if(p.x==ex&&p.y==ey) {return;}

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

            np.x=p.x+dir[i][0];

            np.y=p.y+dir[i][1];

            if(np.x>0 && np.x<=n && np.y>0 && np.y<=m && path[np.x][np.y]>path[p.x][p.y]+1 && word[np.x][np.y]!='X')

            {

                path[np.x][np.y]=path[p.x][p.y]+1;

                que.push(np);

            }

        }

    }

}

int main()

{

#ifdef LOCAL

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

	//freopen("out.txt","w",stdout);

#endif

	ios_base::sync_with_stdio(0);



	while(scanf("%d%d%d",&n,&m,&t)!=EOF){

        ans=0;

        CLR(word);

        for(int i=1;i<=n;scanf("%s",word[i++]+1));

        for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(word[i][j]=='S') word[i][j]='0';

        for(int k=0;k<t;k++){

            char tp=k+'0';

            for(int i=1;i<=n;i++) for(int j=1;j<=m;j++){

                if(word[i][j]==tp) sx=i,sy=j;

                if(word[i][j]==tp+1) ex=i,ey=j;

            }

            bfs();

            ans+=path[ex][ey];

            //printf("%d\t",path[ex][ey]);

        }

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

	}



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 6.144516061977945x):
#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;







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

// Method:    bfs

// FullName:  bfs

// Access:    public

// Returns:   int

// Qualifier:

// Parameter: const int & sx 起点x

// Parameter: const int & sy 起点y

// Parameter: const int & gx ?点x

// Parameter: const int & gy ?点y

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

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;

}

///////////////////////////End Sub//////////////////////////////////