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 namespace std;
#define map Map
 
int H,W,K;
char map[1000][1000];
int dst[1000][1000];

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

void bfs(int sx, int sy){
	typedef pair<int,int> pii;
	queue<pii> q;
	q.push(pii{sx, sy});
	dst[sx][sy]=1;
	while(q.size()){
		int x, y; tie(x, y) = q.front(); q.pop();
		if(dst[x][y] == K+1) continue;
		for(int d=0; d<4; ++d){
			int nx=x+dx[d], ny=y+dy[d];
			if(nx>=0 && ny>=0 && nx<H && ny<W && map[nx][ny]!='#' && !dst[nx][ny]){
				dst[nx][ny] = dst[x][y]+1;
				q.push(pii{nx, ny});
			}
		}
	}
}

int main()
{
	cin >> H>>W>>K;
	for (int i = 0; i < H; i++)
	{
		cin >> map[i];
	}
	int sx, sy;
	for (int i = 0; i < H; i++)
	{
		for (int w = 0; w < W; w++)
		{
			if (map[i][w] == 'S') {
				sx = i; sy = w; map[i][w] = '.';
			}
		}
	}
	bfs(sx, sy);
 
	int res = 1e9;
	for (int i = 0; i < H; i++)
	{
		for (int w = 0; w < W; w++)
		{
			if (map[i][w] != '#' && dst[i][w]) {
				res = min(res, 1 + (min({ i,H - i - 1,w,W - w - 1 })+K-1)/K);
			}
		}
	}
	cout << res << endl;
 
	return 0;
}