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:
// B.

#include <iostream>
#include <algorithm>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

typedef pair<int, int> II;

void update_r(int h, int w, int x, int y, int &r) {
	r = min(r, x);
	r = min(r, y);
	r = min(r, w - 1 - x);
	r = min(r, h - 1 - y);
}

int main(int argc, char *argv[])
{
	int h, w, k, sx, sy, r = 1 << 30;
	cin >> h >> w >> k;
	string b[1000];
	for (int i = 0; i < h; ++i) {
		cin >> b[i];
		int pos = b[i].find('S');
		if (pos >= 0) {
			sx = pos, sy = i;
		}
	}
	update_r(h, w, sx, sy, r);
	vector<II> v;
	v.push_back(II(sx, sy));
	for (int i = 0; i < k; ++i) {
		vector<II> next;
		for (auto &xy : v) {
			int dx[] = {-1,1,0,0};
			int dy[] = {0,0,-1,1};
			for (int d = 0; d < 4; ++d) {
				int x = xy.first + dx[d], y = xy.second + dy[d];
				if (x >= 0 && x < w && y >= 0 && y < h && b[y][x] == '.') {
					b[y][x] = '?';
					next.push_back(II(x, y));
					update_r(h, w, x, y, r);
				}
			}
		}
		v = next;
	}
	int ans = 1 + (r + k - 1) / k;
	cout << ans << endl;
	return 0;
}