// Slower C++ solution:
#include <iostream>

#include <queue>

#include <string.h>

using namespace std;

typedef pair<int, int> P;

int sx,sy,H,W,N;

char map1[1002][1002];

int step[1002][1002];

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

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

void find1(int g){

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

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

        if(map1[i][k] == 'S' || map1[i][k] == (g+48)){

        if(map1[i][k] == 'S') map1[i][k] ='.';

        sx = i; sy = k;

        return ;

    }

}



int bfs(int n)

{

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

    queue <P> que;

    que.push(P (sx,sy));

    step[sx][sy] = 0;

    while(que.size()){

        int x = que.front().first, y = que.front().second;

        que.pop();

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

            int xn = x + dx[i], yn = y + dy[i];

            if(xn >= 0 && xn < W && yn >= 0 && yn < H && map1[xn][yn] != 'X' && step[xn][yn] == -1)

                {

                    step[xn][yn] = step[x][y] + 1;

                    if(map1[xn][yn] == (n+48))

                        {

                            return step[xn][yn];

                        }

                    que.push(P(xn, yn));

                }

        }

    }

return 0;

}



int main()

{

    cin >> W >> H >> N;

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

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

        cin >> map1[i][k];

    int step = 0;

    find1(0);

    step += bfs(1);

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

    {



        find1(i);

        step += bfs(i+1);

    }

    cout << step << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.911894571012502x):
#include<iostream>

#include<cstdio>

#include<vector>

#include<cmath>

#include<map>

#include<set>

#include<queue>

#include<cstring>

#include<algorithm>

using namespace std;

const int maxn = 1000 + 8;

#define inf 0x3f3f3f3f



struct node

{

    int x,y;

    node(int x = 0,int y = 0):x(x),y(y){}

}s,current,st,nd;



int h,w,n;

int step[maxn][maxn];

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

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

string Map[maxn];

int sum = 0;



int bfs()

{

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

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

            step[i][j] = inf;

        }

    }

    queue<node> q;

    step[st.x][st.y] = 0;

    q.push(st);

    while(!q.empty())

    {

        current = q.front();

        q.pop();

        if(current.x == nd.x && current.y == nd.y) break;

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

        {

            s.x = current.x + dx[i];

            s.y = current.y + dy[i];

            if(s.x >= 0 && s.x < h && s.y >= 0 && s.y < w && Map[s.x][s.y] != 'X' && step[s.x][s.y] == inf)

            {

                step[s.x][s.y] = step[current.x][current.y] + 1;

                q.push(s);

            }

        }

    }

    return step[nd.x][nd.y];

}

int main()

{

    while(cin >> h >> w >> n)

    {

        sum = 0;

        for(int i = 0 ; i < h; ++i) cin >> Map[i];

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

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

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

                    st.x = i;

                    st.y = j;

                }

            }

        }



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

        {

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

            {

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

                {

                    if(Map[j][k] == '1' + i)

                    {

                        nd.x = j; nd.y = k;

                        sum += bfs();

                        st.x = j; st.y = k;

                    }

                }

            }

        }

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

    }

    return 0;

}



/*

3 3 1

S..

...

..1

4 5 2

.X..1

....X

.XX.S

.2.X.

10 10 9

.X...X.S.X

6..5X..X1X

...XXXX..X

X..9X...X.

8.X2X..X3X

...XX.X4..

XX....7X..

X..X..XX..

X...X.XX..

..X.......

*/