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;

#define iter(i, n) for (int i = 1; i <= n; ++i)
#define iter0(i, n) for (int i = 0; i < n; ++i)

typedef pair<int, int> pii;

#define fi first
#define se second

const int NR = 1010;

int n, m, K, dt[NR][NR];
char s[NR][NR];
pii q[NR * NR];

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

bool valid(int x, int y) {
	return 1 <= x && x <= n && 1 <= y && y <= m && s[x][y] == '.';
}

int mind(int x, int y) {
	return min(min(x - 1, n - x), min(y - 1, m - y));
}

int main() {
//	freopen("14C.in", "r", stdin);
	scanf("%d%d%d", &n, &m, &K);
	int sx, sy;
	iter(i, n) {
		scanf("%s", s[i] + 1);
		iter(j, m) { if (s[i][j] == 'S') sx = i, sy = j; else dt[i][j] = -1; }
	}

	int h, t;

	q[h = t = 1] = pii(sx, sy);

	int ans = 1 + (mind(sx, sy) + K - 1) / K;

	while (h <= t) {
		pii u = q[h++]; int x = u.fi, y = u.se;
//		printf("dt[%d][%d]=%d\n", x, y, dt[x][y]);
		iter0(d, 4) {
			int _x = x + dx[d], _y = y + dy[d];
			if (valid(_x, _y) && dt[_x][_y] == -1) dt[_x][_y] = dt[x][y] + 1, q[++t] = pii(_x, _y);
		}

		if (dt[x][y] <= K) ans = min(ans, 1 + (mind(x, y) + K - 1) / K);
	}

	printf("%d\n", ans);
	return 0;
}