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;
const int N = 800 + 10;
int n, m, k, dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, mark[N][N], dis[N][N], ans = N + 1;
string s[N];
pair<int, int> st;
bool isval(int x, int y) {
	if (x < n && x >= 0 && y < m && y >= 0 && s[x][y] == '.')
		return true;
	return false;
}
void bfs(int sx, int sy) {
	mark[sx][sy] = 1;
	queue<pair<int, int> > q;
	q.push({sx, sy});
	dis[sx][sy] = 0;
	while (!q.empty()) {
		auto p = q.front();
		q.pop();
		int x = p.first, y = p.second;
		for (int i = 0; i < 4; i++) {
			int xx = x + dx[i], yy = y + dy[i];
			if (isval(xx, yy) && !mark[xx][yy]) {
				mark[xx][yy] = 1;
				dis[xx][yy] = dis[x][y] + 1;
				q.push({xx, yy});
			}
		}
	}
}
int main() {
	cin >> n >> m >> k;
	for (int i = 0; i < n; i++) {
		cin >> s[i];
		for (int j = 0; j < m; j++)
			if (s[i][j] == 'S')
				st = {i, j};
	}
	memset(dis, 127, sizeof(dis));
	bfs(st.first, st.second);
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			if (dis[i][j] <= k) {
				ans = min({ans, (i + k - 1) / k, (j + k - 1) / k, (n - i - 1 + k - 1) / k, (m - j - 1 + k - 1) / k});
			}	
		}
	}
	cout << ans + 1;
	return 0;
}