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;

typedef long long ll;
typedef pair<int, int> pii;

const int N = 1e3;
string s[N];

const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int dis[N][N], h, w, k;

void bfs(int sx, int sy){
	memset(dis, -1, sizeof dis);
	dis[sx][sy] = 0;
	queue<pii> q;
	q.push(pii(sx, sy));

	while(q.size()){
		int r = q.front().first, c = q.front().second;
		q.pop();

		for (int i = 0; i < 4; i++){
			int nr = r + dx[i], nc = c + dy[i];
			if (nr >= 0 && nr < h && nc >= 0 && nc < w && dis[nr][nc] == -1 && s[nr][nc] == '.'){
				dis[nr][nc] = dis[r][c] + 1;
				q.push(pii(nr, nc));
			}
		}
	}
}

int main(){
	ios::sync_with_stdio(0);
	cout.tie(0); cin.tie(0);

	cin >> h >> w >> k;

	int sx, sy;
	for (int i = 0; i < h; i++){
		cin >> s[i];
		for (int j = 0; j < w; j++)
			if (s[i][j] == 'S') sx = i, sy = j, s[i][j] = '.';
	}

	bfs(sx, sy);

	int ans = 1e9;
	for (int i = 0; i < h; i++)
		for (int j = 0; j < w; j++) if (dis[i][j] <= k && dis[i][j] != -1){
			int t = min({i, j, h - i - 1, w - j - 1});
			//cerr << i << ", " << j << " : " << t << endl;
			ans = min(ans, (t + k - 1) / k + 1);
		}

	cout << ans << endl;
}