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 <cstdio>
#include <algorithm>

const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int MN = 805;

int N, M, K, sx, sy;
char A[MN][MN];
int qx[MN * MN], qy[MN * MN], ql, qr;
int dis[MN][MN];
int Ans;

int main() {
	scanf("%d%d%d", &N, &M, &K), Ans = N;
	for (int i = 1; i <= N; ++i) {
		scanf("%s", A[i] + 1);
		for (int j = 1; j <= M; ++j) {
			dis[i][j] = -1;
			if (A[i][j] == 'S') sx = i, sy = j;
		}
	}
	dis[sx][sy] = 0, ql = qr = 1, qx[1] = sx, qy[1] = sy;
	while (ql <= qr) {
		int x = qx[ql], y = qy[ql]; ++ql;
		int di = std::min({x - 1, N - x, y - 1, M - y});
		Ans = std::min(Ans, 1 + (di + K - 1) / K);
		if (dis[x][y] == K) continue;
		for (int d = 0; d < 4; ++d) {
			int nx = x + dx[d], ny = y + dy[d];
			if (nx < 1 || nx > N || ny < 1 || ny > M || A[nx][ny] == '#' || ~dis[nx][ny]) continue;
			dis[nx][ny] = dis[x][y] + 1;
			++qr, qx[qr] = nx, qy[qr] = ny;
		}
	}
	printf("%d\n", Ans);
	return 0;
}