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 <queue>
using namespace std;

int H,W,K;
int sx,sy;
int dx[4] = {-1,1,0,0},dy[4] = {0,0,-1,1};
char A[810][810];
int B[810][810] = {{0}},visited[810][810] = {0};
bool in(int x, int y){
	return 0<=x && x<=H-1 && 0<=y && y<=W-1;
}

void bfs(void){
	visited[sx][sy] = 1;
	B[sx][sy] = 0;
	queue<pair<int,int>> Q;
	Q.push({sx,sy});
	while(!Q.empty()){
		pair<int,int> p = Q.front();
		Q.pop();
		int x = p.first,y = p.second;
		for(int i=0;i<4;i++){
			int nx = x+dx[i],ny = y+dy[i];
			if(in(nx,ny) && visited[nx][ny]==0 && B[x][y]<K && A[nx][ny]=='.'){
				B[nx][ny] = B[x][y]+1;
				Q.push({nx,ny});
				visited[nx][ny] = 1;
			}
		}
	}
}

int main(){
	cin >> H >> W >> K;
	for(int i=0;i<H;i++){
		for(int j=0;j<W;j++){
			cin >> A[i][j];
			if(A[i][j]=='S'){
				sx = i; sy = j;
			}
		}
	}
	bfs();
	B[sx][sy] = 1;
	int ans = 1e9;
	for(int i=0;i<H;i++){
		for(int j=0;j<W;j++){
			if(B[i][j]!=0){
				int t = min(min(i,H-1-i),min(j,W-j-1));
				ans = min(ans,1+(t+K-1)/K);
				if(i==0 || i==H-1 || j==0 || j==W-1) ans = 1;
			}
		}
	}
	cout << ans << endl;
}