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 = 805;

struct point {
	int x, y, d;
};

int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

int n, m, k;
char str[MAXN][MAXN];
int vis[MAXN][MAXN];

int main()
{
	ios::sync_with_stdio(false);
	cin >> n >> m >> k;
	for (int i = 1; i <= n; i++)
		cin >> str[i]+1;
	point rt;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			if (str[i][j] == 'S')
				rt = (point){i, j, 0};
	static queue<point> que;
	int ans = INT_MAX;
	auto div_up = [](int a, int b){ return a/b+(a%b!=0); };
	vis[rt.x][rt.y] = 1, que.push(rt);
	while (!que.empty()) {
		point nd = que.front(); que.pop();
		if (nd.d > k) continue;
		ans = min(ans, div_up(min(min(nd.x-1, n-nd.x), min(nd.y-1, m-nd.y)), k)+1);
		for (int i = 0; i < 4; i++) {
			int x = nd.x+dx[i], y = nd.y+dy[i];
			if (x < 1 || x > n || y < 1 || y > m || str[x][y] == '#' || vis[x][y]) continue;
			vis[x][y] = 1;
			que.push((point){x, y, nd.d+1});
		}
	}
	cout << ans << endl;
	return 0;
}