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 maxn = 1000;
char grid[maxn][maxn];
int dist[maxn][maxn];
int main(){
	int n, m, k; cin >> n >> m >> k;
	memset(grid, '#', sizeof(grid));
	for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) cin >> grid[i][j];
	memset(dist, -1, sizeof(dist));
	queue<pair<int,int> > q;
	for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(grid[i][j] == 'S'){
		dist[i][j] = 0;
		q.push({i, j});
	}
	while(!q.empty()){
		int x, y; tie(x, y) = q.front(); q.pop();
		for(int dx = -1; dx <= 1; dx++) for(int dy = -1; dy <= 1; dy++){
			if(abs(dx) + abs(dy) == 1){
				int nx = x + dx, ny = y + dy;
				if(grid[nx][ny] != '#' && dist[nx][ny] == -1){
					dist[nx][ny] = dist[x][y] + 1;
					q.push({nx, ny});
				}
			}
		}
	}
	int ans = 1<<30;
	for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(dist[i][j] != -1 && dist[i][j] <= k){
		ans = min(ans, 1 + (min({i - 1, n - i, j - 1, m - j}) + k - 1) / k);
	}
	cout << ans << endl;
	return 0;
}