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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 100000000



int main(){
	
	int H,W,K;
	cin>>H>>W>>K;
	
	vector<string> S(H);
	for(int i=0;i<H;i++){
		cin>>S[i];
	}
	
	int sx,sy;
	
	for(int i=0;i<H;i++){
		for(int j=0;j<W;j++){
			if(S[i][j]=='S'){
				sx=j;
				sy=i;
				break;
			}
		}
	}
	
	vector<vector<int>> dis(H,vector<int>(W,Inf));
	dis[sy][sx]=0;
	queue<pair<int,int>> Q;
	Q.emplace(sy,sx);
	
	while(Q.size()!=0){
		int y = Q.front().first;
		int x = Q.front().second;
		Q.pop();
		for(int i=0;i<4;i++){
			int yy = y;
			int xx = x;
			if(i==0)yy++;
			if(i==1)yy--;
			if(i==2)xx++;
			if(i==3)xx--;
			if(yy<0||yy>=H||xx<0||xx>=W)continue;
			if(S[yy][xx]=='#')continue;
			if(dis[yy][xx]!=Inf)continue;
			dis[yy][xx] = dis[y][x] + 1;
			Q.emplace(yy,xx);
		}
	}
	
	int ans = Inf;
	
	for(int i=0;i<H;i++){
		for(int j=0;j<W;j++){
			if(dis[i][j]>K)continue;
			int a = min({i,j,H-1-i,W-1-j});
			ans = min(ans,(a+K-1)/K);
		}
	}
	cout<<ans+1<<endl;


	return 0;
}