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 <bits/stdc++.h>
using std::cout;
using std::endl;
using std::cin;
constexpr int vx[] = {1, 0, -1, 0}, vy[] = {0, 1, 0, -1};

std::vector<std::string> s;
int h, w, k, sx, sy;

int dist(int x, int y){
	int A = std::min({abs(x - (h - 1)), x, abs(y - (w - 1)), y});
	assert(A >= 0);
	
	return (A + k - 1) / k;
}

int main(){
	cin >> h >> w >> k; s.resize(h);
	for(int i = 0; i < h; i++){
		cin >> s[i];
		
		for(int j = 0; j < w; j++){
			if(s[i][j] != 'S') continue;
			sx = i; sy = j;
			break;
		}
	}
	
	std::queue<std::pair<int, int>> qu; qu.push({sx, sy});
	std::vector<std::vector<int>> min_cost(h, std::vector<int>(w, 1e9));
	min_cost[sx][sy] = 0;
	while(!qu.empty()){
		auto p = qu.front(); qu.pop();
		
		for(int i = 0; i < 4; i++){
			int nx = p.first + vy[i], ny = p.second + vx[i];
			if(nx<0||ny<0||nx>=h||ny>=w||s[nx][ny] == '#') continue;
			if(min_cost[p.first][p.second] + 1 >= min_cost[nx][ny]) continue;
			min_cost[nx][ny] = min_cost[p.first][p.second] + 1;
			qu.push({nx, ny});
		}
	}
	
	int ans = 1e9;
	for(int i = 0; i < h; i++)
		for(int j = 0; j < w; j++)
			if(min_cost[i][j] <= k) ans = std::min(ans, dist(i, j) + 1);
			
	cout << ans << endl;
	return 0;
}