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
#include <bits/stdc++.h>
#define mp make_pair
using namespace std;
string s[1000];
int dis[805][805];
int vis[805][805];
int main(){
	int n, m, k;
	cin >> n >> m >> k;
	int sx, sy;
	for(int i = 0; i < n; i++){
		cin >> s[i];
		for(int j = 0; j < m; j++){
			if(s[i][j] == 'S')sx = i, sy = j;
			dis[i][j] = 1e9;
			vis[i][j] = 0;
		}
	}
	int ans = 1e9;
	queue<pair<int,int> > q;
	q.push(make_pair(sx, sy));
	dis[sx][sy] = 0;
	int xx[4] = {0, 0, 1, -1};
	int yy[4] = {1, -1, 0 ,0};
	while(!q.empty()){
		int x = q.front().first, y = q.front().second;
		q.pop();
		if(vis[x][y] != 0)continue;
		if(dis[x][y] > k)continue;
		ans = min(ans, 1 + (x+k-1)/k);
		ans = min(ans, 1 + (y+k-1)/k);
		ans = min(ans, 1 + (n-x+k-2)/k);
		ans = min(ans, 1 + (m-y+k-2)/k);
		for(int i = 0; i < 4; i++){
			int xxx = x + xx[i], yyy = y + yy[i];
			if(xxx >= 0 && xxx < n && yyy >= 0 && yyy < m && s[xxx][yyy] != '#' && dis[xxx][yyy] > dis[x][y]+1 && vis[xxx][yyy] == 0){
				dis[xxx][yyy] = dis[x][y]+1;
				q.push(mp(xxx,yyy));
			}
		}
	}
	cout << ans << endl;
}