// Slower C++ solution:
#include"stdio.h"

#include"stdlib.h"

#include"string.h"

#include"algorithm"

#include"queue"

using namespace std;

struct mmm

{

	int x,y;

}vv;

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

char xx[1005][1005];

int bb[11][2];

queue <mmm> qq;

int vis[1005][1005];

int bfs(char t)

{

	int sum = 0;

	int pl = 0;

	while(!qq.empty())

	{

		vv = qq.front();

		int x0 = vv.x;

		int y0 = vv.y;

		//printf("%d %d %d\n",x0,y0,vis[x0][y0]);

		qq.pop();

		if(xx[x0][y0] == t && !pl)

		{

			sum = vis[x0][y0];

			pl = 1;

		}

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

		{

			int xt = x0 + dir[i][0];

			int yt = y0 + dir[i][1];

			if(xx[xt][yt] != '#' && xx[xt][yt] != 'X' && vis[xt][yt] == -1)

			{

				vis[xt][yt] = vis[x0][y0] + 1;

				vv.x = xt;

				vv.y = yt;

				qq.push(vv);

			}

		}

	}

	return sum;

}

int main()

{

	int a,b,c,d,x1,y1,sum;

	while(~scanf("%d %d %d",&a,&b,&c))

	{

	//	if(a == 0 && b == 0)

	//	break;

		memset(xx,'#',sizeof(xx));

		getchar();

		for(int i = 1;i <= a;i++)

		{

			for(int j = 1;j <= b;j++)

			{

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

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

				{

					x1 = i;

					y1 = j;

				}

				else if(xx[i][j] != 'X')

				{

					bb[xx[i][j] - '0'][0] = i;

					bb[xx[i][j] - '0'][1] = j;

				}

			}

			getchar();

		}

		bb[0][0] = x1;

		bb[0][1] = y1;

		sum = 0;

		for(int i = 1;i <= c;i++)

		{

			vv.x = bb[i-1][0];

			vv.y = bb[i-1][1];

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

			vis[vv.x][vv.y] = 0;

			qq.push(vv);

			sum += bfs(i + '0');

		}

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

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 6.043821528890767x):
#include<iostream>

#include<iomanip>

#include<cstdlib>

#include<math.h>

#include<algorithm>

#include<string>

#include<vector>

#include<list>

#include<queue>

#include<stack>

#include<deque>

#include<set>

#include<map>

#define REP(i, N) for(ll i = 0; i < N; ++i)

#define FOR(i, a, b) for(ll i = a; i < b; ++i)

#define ALL(a) (a).begin(),(a).end()

#define pb push_back

#define INF 1000000000

#define MOD 1000000007

using namespace std;

typedef long long ll;

typedef pair<ll, ll> P;

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

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

int qx[8] = {1, 1, 0, -1, -1, -1, 0, 1};

int qy[8] = {0, 1, 1, 1, 0, -1, -1, -1};



int main(void) {

	ll h, w, n;

	cin>>h>>w>>n;

	vector<string> a(h);

	vector<vector<ll> > dist(h, vector<ll>(w, INF));

	ll time = 0;

	P sp;

	REP(i, h) cin>>a[i];

	bool det = false;

	REP(i, h) {

		REP(j, w) {

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

				sp.first = i;

				sp.second = j;

				det = true;

				break;

			}

		}

		if(det) break;

	}

	REP(i, n) {

		queue<P> c;

		c.push(sp);

		dist[sp.first][sp.second] = 0;

		while(true) {

			P cp = c.front();

			c.pop();

			if(a[cp.first][cp.second] - '1' == i) {

				time += dist[cp.first][cp.second];

				sp = cp;

				break;

			}

			REP(j, 4) {

				ll nx = cp.first + dx[j];

				ll ny = cp.second + dy[j];

				if(nx >= 0 && nx < h && ny >= 0 && ny < w && dist[nx][ny] == INF && (a[nx][ny] == 'S' || a[nx][ny] == '.' || (a[nx][ny] >= '1' && a[nx][ny] <= '9'))) {

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

					dist[nx][ny] = dist[cp.first][cp.second] + 1;

				}

			}

		}

		REP(j, h) {

			REP(k, w) {

				dist[j][k] = INF;

			}

		}

	}

	cout<<time<<endl;

}