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"
using namespace std;
const int N = 805;
const int dx[4] = {0 , 1 , 0 , -1};
const int dy[4] = {1 , 0 , -1 , 0};
int n , m , k;
char str[N][N];
int dist[N][N];
pair < int , int > que[N * N];
int ql , qr;
int main(){
	scanf("%d %d %d" , &n , &m , &k);
	for(int i = 1 ; i <= n ; ++i){
		for(int j = 1 ; j <= m ; ++j){
			dist[i][j] = 1e9;
		}
	}
	for(int i = 1 ; i <= n ; ++i){
		scanf("%s" , str[i] + 1);
		for(int j = 1 ; j <= m ; ++j){
			if(str[i][j] == 'S'){
				que[qr++] = make_pair(i , j);
				dist[i][j] = 0;
			}
		}
	}
	while(ql < qr){
		int x = que[ql].first;
		int y = que[ql].second;
		++ql;
		for(int i = 0 ; i < 4 ; ++i){
			int nx = x + dx[i];
			int ny = y + dy[i];
			if(nx < 1 || ny < 1 || nx > n || ny > m){
				continue;
			}
			if(str[nx][ny] == '#'){
				continue;
			}
			if(dist[nx][ny] > dist[x][y] + 1){
				dist[nx][ny] = dist[x][y] + 1;
				que[qr++] = {nx , ny};
			}
		}
	}
	int ans = 1e9;
	for(int i = 1 ; i <= n ; ++i){
		for(int j = 1 ; j <= m ; ++j){
			if(dist[i][j] <= k){
				ans = min(ans , 1 + ((min(min(i - 1 , n - i) , min(j - 1 , m - j)) + k - 1) / k));
			}
		}
	}
	printf("%d\n" , ans);
}