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
#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
int main(){
	ll h, w, k; cin >> h >> w >> k;
	string s[h];
	for(ll i = 0; i < h; ++i) cin >> s[i];
	set<pair<ll, ll>> se;
	pair<ll, ll> start;
	for(ll i = 0; i < h; ++i){
		for(ll j = 0; j < w; ++j){
			if(s[i][j] == 'S') start = pair<ll, ll>(i, j);
		}
	}
	se.insert(start);
	queue<pair<pair<ll, ll>, ll>> q;
	q.push(pair<pair<ll, ll>, ll>(start, 0));
	ll dx[] = {1, 0, -1, 0};
	ll dy[] = {0, 1, 0, -1};
	while(q.size()){
		pair<pair<ll, ll>, ll> now = q.front();
		q.pop();
		for(ll i = 0; i < 4; ++i){
			pair<ll, ll> next = pair<ll, ll>(now.first.first + dx[i], now.first.second + dy[i]);
			if(!(0 <= next.first && next.first < h && 0 <= next.second && next.second < w)) continue;
			if(s[next.first][next.second] == '#') continue;
			if(se.find(next) != se.end()) continue;
			if(now.second + 1 > k) continue;
			se.insert(next);
			q.push(pair<pair<ll, ll>, ll>(next, now.second + 1));
		}
	}

	ll ans = 1145148931919;
	for(auto ite = se.begin(); ite != se.end(); ++ite) ans = min(ans, min(min(ite->first, (h - 1) - ite->first), min(ite->second, (w - 1) - ite->second)));
	cout << 1 + ans / k + (ans % k > 0) << endl;
	return 0;
}