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;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

char g[800][801];
int dx[4] = { 1, -1, 0, 0 }, dy[4] = { 0, 0, 1, -1 };

int main(){
	int n, m, k;
	scanf("%d%d%d", &n, &m, &k);
	f(i, 0, n)scanf("%s", g[i]);
	pair<int, int> s;
	f(i, 0, n)f(j, 0, m)if (g[i][j] == 'S') { s = make_pair(i, j); g[i][j] = 1; break; }
	queue<pair<int, pair<int, int> > > q;
	q.push(make_pair(0, s));
	while (!q.empty()){
		int a = q.front().second.first, b = q.front().second.second, c = q.front().first;
		q.pop();
		if (c == k)continue;
		f(i, 0, 4){
			int x = a + dx[i], y = b + dy[i];
			if (x == -1 || x == n || y == -1 || y == m || g[x][y] != '.')continue;
			g[x][y] = 1;
			q.push(make_pair(c + 1, make_pair(x, y)));
		}
	}
	int an = 1e9;
	f(i, 0, n)f(j, 0, m)if (g[i][j] == 1)an = min(an, min(i, min(j, min(n - (i + 1), m - (j + 1)))));
	an = (an + k - 1) / k;
	printf("%d\n", an + 1);
}