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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define ran 888

const int go[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};

int n, m, k;
char s[ran][ran];
int d[ran][ran];
int q[ran*ran*2], l, r;

void bfs() {
	while(l<r) {
		int a = q[l++], b = q[l++];
		int dist = d[a][b];
		if(dist == k) continue;
		for(int i=0;i<4;i++) {
			int x = a+go[i][0], y = b+go[i][1];
			if(x<0 || x>=n || y<0 || y>=m) continue;
			if(s[x][y]=='#' || d[x][y]!=-1) continue;
			
			q[r++] = x;
			q[r++] = y;
			d[x][y] = dist + 1;
		}
	}
}

int calc(int x, int y) {
	int res = x+y;
	res = min(res, (x+k-1)/k);
	res = min(res, ((n-1-x) + (k-1))/k);
	res = min(res, (y+k-1)/k);
	res = min(res, ((m-1-y) + (k-1))/k);
	return res;
}

int main() {
	scanf("%d%d%d", &n, &m, &k);
	memset(d, -1, sizeof d);
	l=r=0;
	for(int i=0;i<n;i++) {
		scanf("%s", s[i]);
		for(int j=0;j<=m;j++)
			if(s[i][j]=='S') {
				q[r++] = i; q[r++] = j; d[i][j] = 0;
				s[i][j] = '.';
			}
	}
	bfs();
	
	int res = n+m;
	for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
			if(d[i][j]!=-1)
				res = min(res, calc(i, j) + 1);
	printf("%d\n", res);
	
	return 0;
}