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>
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define EPS (1e-10)
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int, int>P;

char a[800][801];
int d[800][800];
int dx[]{ 1,-1,0,0 }, dy[]{ 0,0,1,-1 };
signed main() {
	int h, w, k; scanf("%d%d%d", &h, &w, &k);
	int sx, sy;
	rep(i, h) {
		scanf("%s", a[i]);
		rep(j, w) {
			if (a[i][j] == 'S')sx = i, sy = j;
		}
	}
	queue<P>que;
	memset(d, -1, sizeof(d));
	d[sx][sy] = 0; que.push(P(sx, sy));
	while (!que.empty()) {
		P p = que.front(); que.pop();
		rep(i, 4) {
			int nx = p.first + dx[i], ny = p.second + dy[i];
			if (0 <= nx&&nx < h && 0 <= ny&&ny < w&&d[nx][ny] == -1 && a[nx][ny] != '#') {
				d[nx][ny] = d[p.first][p.second] + 1;
				if (d[nx][ny] < k)que.push(P(nx, ny));
			}
		}
	}
	int Min = INF;
	rep(i, h)rep(j, w) {
		if (d[i][j] != -1) {
			int c = min({ i, h - i - 1,j, w - j - 1 });
			Min = min(Min, c / k + (c%k > 0) + 1);
		}
	}
	printf("%d\n", Min);
}