// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

const int MAX = 1024;



int h, w, n, maze[MAX][MAX];

int b[15];

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

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

int d[MAX][MAX];

int ans = 0;



int readchar() {

    for(;;) {

        int ch = getchar();

        if(ch != '\n' && ch != '\r') return ch;

    }

}



int bfs(int x, int y, int x2, int y2)

{

    //cout << x << y << x2 << y2 << endl;



    queue<int> q;

    q.push(x * 10000 + y);

    while(!q.empty()) {

        int v = q.front();

        q.pop();

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

            int x1 = v / 10000 + r[i];

            int y1 = v % 10000 + z[i];

            if(x1 == x2 && y1 == y2) {

                d[x1][y1] = d[v / 10000][v % 10000] + 1;

                return 0;

            }

            if(x1 >= 0 && y1 >= 0 && x1 < h && y1 < w && maze[x1][y1] != 'X' && d[x1][y1] == -1) {

                 d[x1][y1] = d[v / 10000][v % 10000] + 1;

                 q.push(x1 * 10000 + y1);

            }

        }

    }

    return 0;

}

int main()

{

    cin >> h >> w >> n;

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

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

             maze[i][j] = readchar();

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

            b[0] = i * 10000 + j;

        }

        if(maze[i][j] != 'S' && maze[i][j] != 'X' && maze[i][j] != '.') {

            b[maze[i][j] - '0'] = i * 10000 + j;

        }

    }

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

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

        d[b[i] / 10000][b[i] % 10000] = 0;

        bfs(b[i] / 10000, b[i] % 10000, b[i + 1] / 10000, b[i + 1] % 10000);

        ans += d[b[i + 1] / 10000][ b[i + 1] % 10000];

    }

    cout << ans << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.244750028033361x):
/* 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;

}