// 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: 4.921431651504528x):
#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.......

*/