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 DX[] = {-1, 1, 0, 0};
const int DY[] = {0, 0, -1, 1};
const int N = 800;
bool visited[N][N];
string s[N];

int main() {
	ios::sync_with_stdio(false);
	int m, n, k; cin >> m >> n >> k;
	for (int i = 0; i < m; ++i) cin >> s[i];
	int sx = -1, sy = -1;
	for (int i = 0; i < m; ++i) {
		for (int j = 0; j < n; ++j) {
			if (s[i][j] == 'S') {
				sx = i;
				sy = j;
			}
		}
	}
	queue<pair<int, int> > q;
	q.emplace(sx, sy);
	visited[sx][sy] = true;
	for (int step = 0; step < k; ++step) {
		for (int zzz = q.size(); zzz > 0; --zzz) {
			int x = q.front().first, y = q.front().second;
			q.pop();
			for (int i = 0; i < 4; ++i) {
				int nx = x + DX[i], ny = y + DY[i];
				if (nx >= 0 && nx < m && ny >= 0 && ny < n && !visited[nx][ny] && s[nx][ny] != '#') {
					visited[nx][ny] = true;
					q.emplace(nx, ny);
				}
			}
		}
	}
	int res = 1e9;
	for (int i = 0; i < m; ++i) {
		for (int j = 0; j < n; ++j) if (visited[i][j]) {
			res = min(res, min(min(i, j), min(m - i - 1, n - j - 1)));
		}
	}
	cout << (res + k - 1) / k + 1 << '\n';
	return 0;
}