// Slower C++ solution:
/*

*

*/

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<string>

#include<queue>

#define CLR(ARR,NUM) memset(ARR,NUM,sizeof(ARR))

#define GETNUM(NUM) scanf("%d",&NUM)

#define faster_io() ios_base::sync_with_stdio(false)

using namespace std;

const int MAXN = 1010;

char map[MAXN][MAXN];

bool vis[10][MAXN][MAXN];

struct node{

	int x;

	int y;

	int deep;

	int n;

};

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

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



int main()

{

	CLR(vis,false);

	CLR(map,'X');

	int h,w,n;

	cin >> h >> w >> n;

	getchar();

	int sx,sy;

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

		for(int j=1;j<=w;j++){

			map[i][j]=getchar();

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

				sx=i;

				sy=j;

			}

		}

		getchar();

	}

	node t;

	t.x=sx;

	t.y=sy;

	t.deep=0;

	t.n=0;

	queue<node>q;

	vis[0][sx][sy]=true;

	q.push(t);

	int ans = -1;

	while(!q.empty()){

		node cur=q.front();

		if(cur.n==n){

			ans=cur.deep;

			break;

		}

		q.pop();

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

			int tx=cur.x+dx[i];

			int ty=cur.y+dy[i];

			if(map[tx][ty]!='X'&&!vis[cur.n][tx][ty]){

				node t = cur;

				t.x=tx;

				t.y=ty;

				t.deep++;

				if(map[tx][ty]=='1'+t.n){

					t.n++;

				}

				vis[t.n][tx][ty]=true;

				q.push(t);

			}

		}

	}

	cout << ans << endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 12.73614427391744x):
#include <iostream>

#include <queue>

using namespace std;

int w, h, n;

char map[1024][1024];

// 各点到当前工厂的距?

int d[1024][1024];

const int direction[4][2] = {

	{ -1, 0 },

	{ 1, 0 },

	{ 0, -1 },

	{ 0, 1 },

};



int factory[16][2];

typedef pair<int, int> P;







//************************************

// Method:    bfs

// FullName:  bfs

// Access:    public

// Returns:   int

// Qualifier:

// Parameter: const int & sx 起点x

// Parameter: const int & sy 起点y

// Parameter: const int & gx ?点x

// Parameter: const int & gy ?点y

//************************************

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 + direction[i][0];

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

			// 是否可以移?，并且?点没有???

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

}



///////////////////////////SubMain//////////////////////////////////

int main(int argc, char *argv[])

{



	cin >> h >> w >> n;

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

	{

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

		{

			cin >> map[j][i];

		}

	}



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

	{

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

		{

			if (map[j][i] == 'S')

			{

				factory[0][0] = j;

				factory[0][1] = i;

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

			}

			else if (isdigit(map[j][i]))

			{

				int index = map[j][i] - '0';

				factory[index][0] = j;

				factory[index][1] = i;

			}

		}

	}



	int step = 0;

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

	{

		// 按?序吃遍中?

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

	}



	cout << step << endl;



	return 0;

}

///////////////////////////End Sub//////////////////////////////////