// 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: 5.065554704274273x):
/* AOJ0558 Cheese */



#include <iostream>

#include <queue>

//#include <string.h>



using namespace std;



const int N = 1000;

char grid[N][N];

int w, h, n;



// 各点到当前工厂的距?

int d[N][N];



const int direct[4][2] = {

    { -1, 0 },

    { 1, 0 },

    { 0, -1 },

    { 0, 1 },

};



int factory[10][2];

typedef pair<int, int> P;



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 + direct[i][0];

            int ny = p.second + direct[i][1];

            // 是否可以移?，并且?点没有???

            if (0 <= nx && nx < w && 0 <= ny && ny < h && grid[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];

}



int main()

{



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

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

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

                cin >> grid[j][i];

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

                    factory[0][0] = j;

                    factory[0][1] = i;

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

                }

                else if (isdigit(grid[j][i])) {

                    factory[grid[j][i] - '0'][0] = j;

                    factory[grid[j][i] - '0'][1] = i;

                }

            }





        int minstep = 0;

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

            minstep += bfs(factory[i][0], factory[i][1], factory[i + 1][0], factory[i + 1][1]);

        }



        cout << minstep << endl;

    }



    return 0;

}