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;
const int di[4] = {-1, 0, 1, 0}, dj[4] = {0, 1, 0, -1}, MAX = 800, INF = 1e9;

int n, m, k, ans = INF, d[MAX][MAX];
string s[MAX];
pair<int,int> startPos;
inline bool valid(int i, int j) {
	return i >= 0 && i < n && j >= 0 && j < m;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m >> k;
	for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) d[i][j] = INF;
	for(int i = 0; i < n; i++) {
		cin >> s[i];
		for(int j = 0; j < m; j++) if(s[i][j] == 'S') startPos = {i, j};
	}
	queue<pair<int,int> > q;
	q.push(startPos);
	d[startPos.first][startPos.second] = 0;
	while(!q.empty()) {
		int i, j;
		tie(i, j) = q.front();
		q.pop();
		for(int z = 0; z < 4; z++) {
			if(valid(i+di[z], j+dj[z]) and d[i+di[z]][j+dj[z]] == INF and s[i+di[z]][j+dj[z]] != '#'){
				q.emplace(i+di[z], j+dj[z]);
				d[i+di[z]][j+dj[z]] = d[i][j]+1;
			}
		}
	}
	for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) if(d[i][j] <= k) {
		ans = min((min(min(i, n-1-i), min(j, m-1-j))+k-1)/k+1, ans);
	}
	cout << ans;
}