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>
#define fi first
#define se second
#define mp make_pair
using namespace std;

typedef pair <int, int> pii;
typedef pair <pii, int> piii;

char a[810][810];
int ans;
queue <piii> q;
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1};
int n, m, k;

bool IN(int x, int y){return x >= 0 && x < n && y >= 0 && y < m;}

int main(){
	scanf("%d%d%d", &n, &m, &k);
	int x, y;
	for (int i = 0; i < n; i++){
		scanf("%s", a[i]);
		for (int j = 0; j < m; j++){
			if (a[i][j] == 'S') x = i, y = j;
		}
	}
	q.push(mp(mp(x, y), k)), a[x][y] = '#';
	int ans = 0x3f3f3f3f;
	while (!q.empty()){
		piii t = q.front(); q.pop();
		int x = t.fi.fi, y = t.fi.se, cnt = t.se;
		ans = min(ans, (min(min(x, y), min(n - x - 1, m - y - 1)) + k - 1) / k);
		if (t.se == 0) continue;
		for (int i = 0; i < 4; i++){
			int xx = x + dir[i][0], yy = y + dir[i][1];
			if (!IN(xx, yy) || a[xx][yy] != '.') continue;
			a[xx][yy] = '#';
			q.push(mp(mp(xx, yy), cnt - 1));
		}
	}
	printf("%d\n", ans + 1);
	return 0;
}