// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 2.6402239516822217x):
/*Lucky_Glass*/

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<queue>

using namespace std;

const int MAXN=1000;

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

struct QUEUE{

	int x,y,dep;

	QUEUE(){}

	QUEUE(int _x,int _y,int _dep){x=_x;y=_y;dep=_dep;}

};

int r,c,n;

int goal[10][2];

bool vis[MAXN+5][MAXN+5];

char maz[MAXN+5][MAXN+5];

int BFS(int *beg,int cnt){

	memset(vis,false,sizeof vis);

	queue<QUEUE> que;

	que.push(QUEUE(beg[0],beg[1],0));

	while(!que.empty())

	{

		QUEUE pre=que.front();que.pop();

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

		{

			int X=pre.x+dir[i][0],Y=pre.y+dir[i][1],depth=pre.dep+1;

			if(X<1 || X>r || Y<1 || Y>c || maz[X][Y]=='X' || vis[X][Y]) continue;

			vis[X][Y]=true;

			if(X==goal[cnt][0] && Y==goal[cnt][1])

				return depth;

			que.push(QUEUE(X,Y,depth));

		}

	}

	return -1;

}

int main(){

	scanf("%d%d%d",&r,&c,&n);

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

	{

		scanf("%s",maz[i]+1);

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

		{

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

				goal[maz[i][j]-'0'][0]=i,

				goal[maz[i][j]-'0'][1]=j,

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

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

				goal[0][0]=i,

				goal[0][1]=j,

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

		}

	}

	int ans=0;

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

		ans+=BFS(goal[i-1],i);

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

	return 0;

}//