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 <iostream>
#include <cstdio>
#include <queue>
using namespace std;

int h, w, k, mn=1e9, d[805][805];
char a[805][805];
queue<pair<int,int>> qu;
int dr[4]={1, 0, -1, 0};
int dc[4]={0, 1, 0, -1};

int main() {
	int i, j, r, c, tr, tc;
	cin >> h >> w >> k;
	for(i=0; i<h; i++) {
		cin >> a[i];
		for(j=0; j<w; j++) {
			d[i][j] = -1;
			if(a[i][j]=='S') {
				d[i][j] = 0;
				qu.push({i, j});
			}
		}
	}
	while(!qu.empty()) {
		r = qu.front().first;
		c = qu.front().second;
		qu.pop();
		mn = min(mn, min(r,c));
		mn = min(mn, min(h-1-r, w-1-c));
		if(0<r && r<h-1 && 0<c && c<w-1 && d[r][c]<k) {
			for(i=0; i<4; i++) {
				tr = r + dr[i];
				tc = c + dc[i];
				if(a[tr][tc]!='#' && d[tr][tc]==-1) {
					d[tr][tc] = d[r][c] + 1;
					qu.push({tr, tc});
				}
			}
		}
	}
	cout << (mn+k-1)/k + 1;
	return 0;
}