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

#include<cstdio>

#include<cstring>

#include<queue>

using namespace std;



const int DIRECTSIZE = 4;

struct _direct {

    int drow;

    int dcol;

} direct[DIRECTSIZE] = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};



const int N = 1000;

const int N2 = 10;

char grid[N][N], visited[N][N];

int h, w, n;

struct _factory {

    int row;

	int col;

} factory[N2];



struct _node {

    int row;

	int col;

	int level;

};



int bfs(int n, char goal) {

    memset(visited, 0, sizeof(visited));

    _node start={factory[n-1].row, factory[n-1].col, 0}, temp;

    visited[start.row][start.col] = 1;

    queue<_node> q;

    q.push(start);

    while(!q.empty()) {

        _node front = q.front();

        q.pop();

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

            int nextrow = front.row + direct[i].drow;

            int nextcol = front.col + direct[i].dcol;

            if( 0 <= nextrow && nextrow < h && 0 <= nextcol && nextcol < w) {

                if(grid[nextrow][nextcol] == goal) {

                    return front.level + 1;

                }

				else if(visited[nextrow][nextcol] == 0 && grid[nextrow][nextcol] != 'X') {

                    visited[nextrow][nextcol] = 1;

                    temp.row = nextrow;

                    temp.col = nextcol;

                    temp.level = front.level + 1;

                    q.push(temp);

                }

            }

        }

    }

    return 0;

}



int main() {

    while (~scanf("%d %d %d", &h, &w, &n)){

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

			scanf("%s", grid[i]);

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

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

                    factory[0].row = i;

                    factory[0].col = j;

				}

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

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

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

                }

            }

		}

        int minstep = 0;

        char goal = '1';

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

            minstep += bfs(i, goal);

            goal++;

        }

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

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.9806086195696844x):
#include<iostream>

#include<algorithm>

#include<queue>

const int INF = 100000000;

using namespace std;

typedef pair<int, int> P;

char a[1005][1005];

int N, M;

int sx, sy;

int gx, gy;

int d[1005][1005];

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

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

int bfs()

{

	queue<P> que;

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

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

			d[i][j] = INF;

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

	d[sx][sy] = 0;

	while (que.size())

	{

		P p = que.front();

		que.pop();

		if (p.first == gx&&p.second == gy)

			break;

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

		{

			int nx = p.first + dx[i];

			int ny = p.second + dy[i];

			if (nx >= 0 && nx < N&&ny >= 0 && ny < M&&a[nx][ny] != 'X'&&d[nx][ny] == INF)

			{

				que.push(P(nx, ny));

				d[nx][ny] = d[p.first][p.second] + 1;

			}

		}



	}

	return d[gx][gy];

}

int main(void)

{

	while (cin >> N >> M)

	{

		int n, sum = 0;

		cin >> n;

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

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

				cin >> a[i][j];

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

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

				if (a[j][k] == 'S')

				{

					sx = j;

					sy = k;

				}

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

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

				if (a[j][k] == '1')

				{

					gx = j;

					gy = k;

				}

		sum += bfs();

		for (int i = 0; i < n - 1; i++)

		{

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

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

				{

					if (a[j][k] == (char)(i+49))

					{

						sx = j;

						sy = k;

					}

					if (a[j][k] == (char)(i+50))

					{

						gx = j;

						gy = k;

					}

				}

			sum += bfs();



		}

		cout << sum << endl;

	}

}