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 dir[4][2] = {1,0,-1,0,0,1,0,-1};
pair < int , int > dst[803][803];
char Map[803][803]; int H , W , K;
struct data{
	pair < int , int > val; pair < int , int > pos;
	friend bool operator <(data p , data q){return p.val > q.val;}
}; priority_queue < data > q;

int main(){
	memset(dst , 0x3f , sizeof(dst));
	cin >> H >> W >> K;
	for(int i = 1 ; i <= H ; ++i)
		for(int j = 1 ; j <= W ; ++j){
			cin >> Map[i][j];
			if(Map[i][j] == 'S'){
				dst[i][j] = make_pair(0 , 0);
				q.push((data){make_pair(0 , 0) , make_pair(i , j)});
			}
		}
	while(!q.empty()){
		pair < int , int > t = q.top().pos;
		int x = t.first , y = t.second;
		if(x == 1 || y == 1 || x == H || y == W){cout << dst[x][y].first + 1; return 0;}
		q.pop();
		for(int i = 0 ; i < 4 ; ++i){
			int X = x + dir[i][0] , Y = y + dir[i][1];
			pair < int , int > ndst = dst[x][y].first || Map[X][Y] != '#' ? (dst[x][y].second == K ? make_pair(dst[x][y].first + 1 , 1) : make_pair(dst[x][y].first , dst[x][y].second + 1)) : make_pair(1 , 1);
			if(dst[X][Y] > ndst){dst[X][Y] = ndst; q.push((data){dst[X][Y] , make_pair(X , Y)});}
		}
	}
	return 0;
}