Problem: Takahashi is locked within a building.

This building consists of H×W rooms, arranged in H rows and W columns. We will denote the room at the i-th row and j-th column as (i,j). The state of this room is represented by a character A_{i,j}. If A_{i,j}= `#`, the room is locked and cannot be entered; if A_{i,j}= `.`, the room is not locked and can be freely entered. Takahashi is currently at the room where A_{i,j}= `S`, which can also be freely entered.

Each room in the 1-st row, 1-st column, H-th row or W-th column, has an exit. Each of the other rooms (i,j) is connected to four rooms: (i-1,j), (i+1,j), (i,j-1) and (i,j+1).

Takahashi will use his magic to get out of the building. In one cast, he can do the following:

* Move to an adjacent room at most K times, possibly zero. Here, locked rooms cannot be entered.
* Then, select and unlock at most K locked rooms, possibly zero. Those rooms will remain unlocked from then on.



His objective is to reach a room with an exit. Find the minimum necessary number of casts to do so.

It is guaranteed that Takahashi is initially at a room without an exit.

Constraints

* 3 ≤ H ≤ 800
* 3 ≤ W ≤ 800
* 1 ≤ K ≤ H×W
* Each A_{i,j} is `#` , `.` or `S`.
* There uniquely exists (i,j) such that A_{i,j}= `S`, and it satisfies 2 ≤ i ≤ H-1 and 2 ≤ j ≤ W-1.

Input

Input is given from Standard Input in the following format:


H W K
A_{1,1}A_{1,2}...A_{1,W}
:
A_{H,1}A_{H,2}...A_{H,W}


Output

Print the minimum necessary number of casts.

Examples

Input

3 3 3
#.#
#S.
###


Output

1


Input

3 3 3
.#
S.


Output

1


Input

3 3 3

S#


Output

2


Input

7 7 2


...##
S###
.#.##
.###


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#define INF 1e+9
using namespace std;

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

typedef pair<int,int> P;

int main(){
	int h,w,k,mi = INF;
	bool used[800][800] = {};
	string field[800];
	cin >> h >> w >> k;
	queue<P> que;
	for(int i = 0;i < h;i++){
		cin >> field[i];
		for(int j = 0;j < w;j++){
			if(field[i][j] == 'S'){
				que.push(P(i,j));
			}
		}
	}
	for(int d = 0;d <= k;d++){
		int siz = que.size();
		if(!siz) break;
		for(int i = 0;i < siz;i++){
			P p = que.front();que.pop();
			int x = p.first,y = p.second;
			mi = min({mi,x,y,h - 1 - x,w - 1 - y});
			for(int j = 0;j < 4;j++){
				int nx = x + dx[j],ny = y + dy[j];
				if(nx >= 0 && nx < h && ny >= 0 && ny < w && field[nx][ny] != '#' && !used[nx][ny]){
					used[nx][ny] = true;
					que.push(P(nx,ny));
				}
			}
		}
	}
	if(!mi) cout << 1 << endl;
	else cout << (mi - 1) / k + 2 << endl;
	return 0;
}