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;
const int N = 805;
int dis[N][N];
char g[N][N];

int dx[] = { 0, 0, -1, 1 };
int dy[] = { 1, -1, 0, 0 };
int R, C, K;
vector<int> v;
void bfs(int x, int y) {

	dis[x][y] = 0;
	queue<pair<int, int> > q;
	q.push( { x, y });
	int T = min(x, min(R - x - 1, min(C - y - 1, y)));
	v.push_back((T + K - 1) / K);
	while (!q.empty()) {
		x = q.front().first;
		y = q.front().second;
		q.pop();
		if (dis[x][y] == K) continue;
		for (int i = 0; i < 4; i++) {
			int tx = dx[i] + x;
			int ty = dy[i] + y;
			if (tx < 0 || tx >= R || ty < 0 || ty >= C || dis[tx][ty] != -1
					|| g[tx][ty] != '.') continue;
			dis[tx][ty] = dis[x][y] + 1;
			int T = min(tx, min(R - tx - 1, min(C - ty - 1, ty)));
			v.push_back((T + K - 1) / K);
			q.push( { tx, ty });
		}
	}
}

int main() {

//	freopen("readin.txt", "r", stdin);

	scanf("%d%d%d", &R, &C, &K);
	int sx, sy;
	for (int i = 0; i < R; i++) {
		for (int j = 0; j < C; j++) {
			scanf(" %c", &g[i][j]);
			if (g[i][j] == 'S') {
				sx = i, sy = j;
			}
		}
	}
	memset(dis, -1, sizeof dis);
	bfs(sx, sy);
	sort(v.begin(), v.end());
	cout << v[0] + 1 << endl;

	return 0;
}
