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 <cstdio>
#include <algorithm>

using namespace std;

int Q[1010000][2];
int head, tail;
int D[810][810];
int n, m, K, sx, sy;
char p[801][801];

void Ins(int x, int y, int d) {
	if (D[x][y] <= d || p[x][y] != '.') {
		return;
	}
	D[x][y] = d;
	tail++;
	Q[tail][0] = x;
	Q[tail][1] = y;
}

int main() {
	int i, j, x, y;
	scanf("%d%d%d", &n, &m, &K);
	for (i = 1; i <= n; i++) {
		scanf("%s", p[i] + 1);
		for (j = 1; j <= m; j++) {
			if (p[i][j] == 'S') {
				sx = i;
				sy = j;
				p[i][j] = '.';
			}
			D[i][j] = 1e9;
		}
	}
	Ins(sx, sy, 0);
	while (head < tail) {
		head++;
		x = Q[head][0];
		y = Q[head][1];
		Ins(x + 1, y, D[x][y] + 1);
		Ins(x, y + 1, D[x][y] + 1);
		Ins(x - 1, y, D[x][y] + 1);
		Ins(x, y - 1, D[x][y] + 1);
	}
	int res = 1e9;
	for (i = 1; i <= n; i++) {
		for (j = 1; j <= m; j++) {
			if (D[i][j] <= K) {
				res = min(res, 1 + (min(min(i - 1, n - i), min(j - 1, m - j)) + K - 1) / K);
			}
		}
	}
	printf("%d\n", res);
	return 0;
}