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 = 8e2 + 7;

char t[N][N];
int n, m, k;
int dis[N][N];
deque < pair <int, int> > d;

int main() {
	cin >> n >> m >> k;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++) {
			cin >> t[i][j];
			if (t[i][j] == 'S') {
				dis[i][j] = 1;
				d.push_back({i, j});
			}
		}
	int res = n * m;
	while (d.size()) {
		int x = d.front().first, y = d.front().second;
		d.pop_front();
		if (dis[x][y] - 1 <= k) {
			res = min({res, 1 + (x - 1 + k - 1) / k, 1 + (y - 1 + k - 1) / k, 1 + (n - x + k - 1) / k, 1 + (m - y + k - 1) / k});
		}
		for (int dx = x - 1; dx <= x + 1; dx += 2) {
			if (dx >= 0 && dx <= n && t[dx][y] != '#' && !dis[dx][y])
				dis[dx][y] = dis[x][y] + 1, d.push_back({dx, y});
		}
		for (int dy = y - 1; dy <= y + 1; dy += 2) {
			if (dy >= 0 && dy <= n && t[x][dy] != '#' && !dis[x][dy]) 
				dis[x][dy] = dis[x][y] + 1, d.push_back({x, dy});
		}
	}
	cout << res;
	return 0;
}