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;

int H, W, K, dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
char g[805][805];
bool vis[805][805];
queue< pair< int, pair<int, int> > > Q;

int d(int x) {
	return x / K + (x % K != 0);
}

int main() {
	scanf("%d%d%d", &H, &W, &K);
	for (int i = 0; i < H; i++) scanf("%s", &g[i]);
	int sx, sy;
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			if (g[i][j] == 'S') {
				sx = i;
				sy = j;
			}
		}
	}
	Q.push(make_pair(0, make_pair(sx, sy)));
	vis[sx][sy] = 1;
	int ans = 1000000000;
	while (!Q.empty()) {
		int dis = Q.front().first, x = Q.front().second.first, y = Q.front().second.second;
		Q.pop();
		for (int i = 0; i < 4; i++) {
			int newx = x + dir[i][0], newy = y + dir[i][1];
			if (newx < 0 || newx >= H || newy < 0 || newy >= W) continue;
			int newdis = dis + 1;
			if (newdis > K) continue;
			if (g[newx][newy] == '#') continue;
			if (vis[newx][newy]) continue;
			else vis[newx][newy] = 1;
			Q.push(make_pair(newdis, make_pair(newx, newy)));
		}
	}
	for (int i = 0; i < H; i++) for (int j = 0; j < W; j++) if (vis[i][j]) ans = min(ans, min(min(d(i), d(H - i - 1)), min(d(j), d(W - j - 1))));
	printf("%d\n", ans + 1);
		
}