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 <cmath>
#include <queue>
#include <string>
#include <vector>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
#pragma warning(disable : 4996)
using namespace std;
const int inf = 1012345678;
int dir[] = { 0, 1, 0, -1 };
int H, W, K, sx, sy, d[809][809]; string s[809];
int main() {
	cin >> H >> W >> K;
	for (int i = 0; i < H; i++) cin >> s[i];
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			if (s[i][j] == 'S') sx = i, sy = j;
			d[i][j] = inf;
		}
	}
	queue<pair<int, int> > que; que.push(make_pair(sx, sy)); d[sx][sy] = 0;
	int ret = inf;
	while (!que.empty()) {
		pair<int, int> u = que.front(); que.pop();
		int z = min({ u.first, H - u.first - 1, u.second, W - u.second - 1 });
		ret = min(ret, (z + K - 1) / K + 1);
		int r = d[u.first][u.second];
		if (r < K) {
			for (int i = 0; i < 4; i++) {
				int tx = u.first + dir[i], ty = u.second + dir[i ^ 1];
				if (0 <= tx && tx < H && 0 <= ty && ty < W && d[tx][ty] == inf && s[tx][ty] != '#') {
					d[tx][ty] = r + 1;
					que.push(make_pair(tx, ty));
				}
			}
		}
	}
	printf("%d\n", ret);
	return 0;
}