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;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

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

int H, W, K;
char c[810][810];
int sx, sy;
int d[810][810];

int main() {
	scanf("%d %d %d", &H, &W, &K);
	memset(d, -1, sizeof(d));
	queue<pii> que;

	rep(i, H) {
		scanf("%s", c[i]);
		rep(j, W) {
			if (c[i][j] == 'S') {
				sy = i;
				sx = j;
				c[i][j] = '.';
				que.push(mp(i, j));
				d[i][j] = 0;
			}
		}
	}

	int ret = 100000;

	while (!que.empty()) {
		pii p = que.front(); que.pop();
		int y = p.fi, x = p.se;
		if (d[y][x] > K) break;

		int t = min({y, x, H-1-y, W-1-x});
		ret = min(ret, (t + K  - 1) / K + 1);

		rep(i, 4) {
			int ny = y + dy[i], nx = x + dx[i];
			if (ny >= 0 && ny < H && nx >= 0 && nx < W && c[ny][nx] == '.' && d[ny][nx] == -1) {
				d[ny][nx] = d[y][x] + 1;
				que.push(mp(ny, nx));
			}
		}
	}

	cout << ret << endl;

	return 0;
}