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;

// why am I so weak

int n, m, k;
char mat[855][855];

int d[855][855];

typedef pair<int, int> P;

#define ft first
#define sd second

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

int main() {
	scanf("%d %d %d", &n, &m, &k);

	for (int i = 0; i < n; i++) {
		scanf("%s", mat[i]);
	}

	P t;

	for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) {
		if (mat[i][j] == 'S') {
			t = P(i, j);
		}
	}

	queue<P> que;

	memset(d, -1, sizeof(d));

	d[t.ft][t.sd] = 0;
	que.push(t);

	while (!que.empty()) {
		P p = que.front(); que.pop();

		if (d[p.ft][p.sd] == k) break;

		for (int i = 0; i < 4; i++) {
			int nx = p.ft + dx[i], ny = p.sd + dy[i];

			if (nx >= 0 && nx < n && ny >= 0 && ny < m && mat[nx][ny] == '.' && d[nx][ny] < 0) {
				d[nx][ny] = d[p.ft][p.sd] + 1;
				que.push(P(nx, ny));
			}
		}
	}

	int res = INT_MAX;

	for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) if (~d[i][j]) {
		res = min(res, (min(min(i, n - i - 1), min(j, m - j - 1)) + k - 1) / k + 1);
	}

	printf("%d\n", res);

	return 0;
}