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>

#define x first
#define y second
#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

typedef pair<int, int> PII;

const int N = 810;
const int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};

int n, m, k;
char Map[N][N];
int dis[N][N];

int BFS(PII S){
	queue<PII> q;
	q.push(S);
	int ans = 1e9;
	while(!q.empty()){
		PII o = q.front(); q.pop();
		if(dis[o.x][o.y] > k) break;
		ans = min(ans, (min(min(n - o.x, o.x - 1), min(o.y - 1, m - o.y)) + k - 1) / k);

		For(i, 0, 3){
			int u = o.x + dx[i], v = o.y + dy[i];
			if(Map[u][v] == '.') q.push(PII(u, v)), dis[u][v] = dis[o.x][o.y] + 1, Map[u][v] = '#';
		}
	}
	return ans;
}

int main(){
	
	scanf("%d%d%d", &n, &m, &k);
	For(i, 1, n) scanf("%s", Map[i] + 1);
	PII S;
	For(i, 1, n) For(j, 1, m) if(Map[i][j] == 'S') S = PII(i, j);
	printf("%d\n", BFS(S) + 1);

	return 0;
}