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 <queue>
#include <algorithm>
using namespace std;

bool inside(int x, int y, int xlim, int ylim){
	return 0 <= x && x < xlim && 0 <= y && y < ylim;
}

struct puni{
	int x, y, d;
	puni(int x, int y, int d){
		this->x = x;
		this->y = y;
		this->d = d;
	}
};

int main(){
	constexpr int dd[] = {0, 1, 0, -1, 0};
	int h, w, k;
	cin >> h >> w >> k;
	string s[h];
	for(int i = 0; i < h; i++) cin >> s[i];
	queue<puni> q;
	for(int i = 0; i < h; i++) for(int j = 0; j < w; j++) if(s[i][j] == 'S'){
		s[i][j] = '.';
		q.push(puni(i, j, 0));
	}
	int ans = 1e9, muni[h][w];
	fill(muni[0], muni[h], 1e9);
	while(!q.empty()){
		puni p = q.front();
		q.pop();
		if(muni[p.x][p.y] <= p.d || k < p.d) continue;
		muni[p.x][p.y] = p.d;
		int d = min({p.x, p.y, h - p.x - 1, w - p.y - 1});
		ans = min(ans, (d + k - 1) / k);
		for(int i = 0; i < 4; i++){
			int nx = p.x + dd[i], ny = p.y + dd[i + 1];
			if(inside(nx, ny, h, w) && s[nx][ny] == '.'){
				q.push(puni(nx, ny, p.d + 1));
			}
		}
	}
	cout << ans + 1 << endl;
}