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
#include <bits/stdc++.h>
using namespace std;

struct P { int x, y; };

char A[800][800];
int used[800][800];

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

int main()
{
	int H, W, K;
	P s;
	cin >> H >> W >> K;
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			cin >> A[i][j];
			if (A[i][j] == 'S') {
				s.x = i;
				s.y = j;
			}
		}
	}
	vector<P> q;
	q.push_back(s);
	used[s.x][s.y] = 1;
	int res = min(s.x, min(s.y, min(H - s.x - 1, W - s.y - 1)));
	for (int i = 0; i < K && !q.empty(); i++) {
		auto tmp = q; q.clear();
		for (auto p : tmp) {
			for (int j = 0; j < 4; j++) {
				int tx = p.x + dx[j], ty = p.y + dy[j];
				if (tx >= 0 && tx < H && ty >= 0 && ty < W && A[tx][ty] != '#' && used[tx][ty] == 0) {
					used[tx][ty] = 1;
					q.push_back((P){ tx, ty });
					res = min(res, min(tx, min(ty, min(H - tx - 1, W - ty - 1))));
				}
			}
		}
	}
	cout << ((res + K - 1) / K + 1) << endl;
	return 0;
}
