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>

#define MAXH (810)

int h, w, k, sx, sy;
int ans, dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
char str[MAXH][MAXH];
int dis[MAXH][MAXH];

void bfs(){
	std::queue<std::pair<int, int> > queue;
	memset(dis, -1, sizeof(dis));
	dis[sx][sy] = 0;
	queue.push({sx, sy});
	while(!queue.empty()){
		int x = queue.front().first, y = queue.front().second;
		queue.pop();
		for(int i = 0; i < 4; ++ i){
			int dx = x + dir[i][0], dy = y + dir[i][1];
			if(dx < 1 || dx > h || dy < 1 || dy > w || ~dis[dx][dy] || str[dx][dy] == '#') continue;
			dis[dx][dy] = dis[x][y] + 1;
			if(dis[dx][dy] <= k){
				queue.push({dx, dy});
				ans = std::min(ans, (std::min(std::min(dx - 1, h - dx), std::min(dy - 1, w - dy)) + k - 1) / k + 1);
			}
		}
	}
}

int main(){
	scanf("%d%d%d", &h, &w, &k);
	for(int i = 1; i <= h; ++ i){
		scanf("%s", str[i] + 1);
		for(int j = 1; j <= w; ++ j){
			if(str[i][j] == 'S'){
				sx = i;
				sy = j;
			}
		}
	}
	ans = (std::min(std::min(sx - 1, h - sx), std::min(sy - 1, w - sy)) + k - 1) / k + 1;
	bfs();
	printf("%d\n", ans);
	return 0;
}