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

#include <cmath>

#include <queue>

#include <cstdio>

#include <cstring>

#include <algorithm>



#define INF 0x7f7f7f7f

#define PI acos(-1.0)



#define L(x)    (x) << 1

#define R(x)    (x) << 1 | 1

#define lowbit(x)   (x)&(-x)

#define Mid(l, r)   (l + r) >> 1

#define Min(x, y)   (x) < (y) ? (x) : (y)

#define Max(x, y)   (x) < (y) ? (y) : (x)

#define iabs(x)     (x) < 0 ? -(x) : (x)

#define CLR(arr, val)    memset(arr, val, sizeof(arr))



using namespace std;



char map[1024][1024];

bool vis[1024][1024];

int cur, h, w;

int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};



struct node {

	int x, y, step;

} sta, now;



bool check(int x, int y) {

	if (x >= 0 && x < h && y >= 0 && y < w && !vis[x][y] && map[x][y] != 'X')	return true;

	return false;

}







void bfs(node s) {

	memset(vis, 0, sizeof(vis));



	queue<node> que;

	s.step = 0;

	que.push(s);

	node next;

	while (!que.empty()) {

//		cout << "h\n";

		s = que.front();

//		cout << s.x << ' ' << s.y << ' ' << cur << endl;

		que.pop();

		if (map[s.x][s.y] == cur + '0')	{

			sta = s;

			return ;

//			s.step;

		}



//		cout << "a\n";

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

//			cout << i << endl;

			next.x = s.x + dir[i][0];

			next.y = s.y + dir[i][1];

			next.step = s.step + 1;

			if (check(next.x, next.y)) {

				vis[next.x][next.y] = true;

				que.push(next);

			}



		}

	}

}



int main() {

	int n;

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

//		memset(map, 0, sizeof(map));

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

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

				cin >> map[i][j];

				if (map[i][j] == 'S')	sta.x = i, sta.y = j, map[i][j] = '.';

			}

		}

		sta.step = 0;

		int ans = 0;

		for (cur = 1; cur <= n; cur++) {

			bfs(sta);

			ans += sta.step;

			//cout << ans << endl;

		}

		cout << ans << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.047487418711581x):
#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;



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;

}