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 Int = int_fast64_t;

Int h, w, k;
vector<string> a;
vector<vector<Int>> v;
Int dx[4] = {0, 0, 1, -1};
Int dy[4] = {1, -1, 0, 0};

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> h >> w >> k;
	a.resize(h);
	for(auto &i:a) cin >> i;
	v.resize(h, vector<Int>(w, -1));
	using P = pair<Int, Int>;
	queue<P> q;
	for(Int i=0; i<h; ++i)
		for(Int j=0; j<w; ++j)
			if(a[i][j] == 'S'){
				v[i][j] = k;
				q.push(P(i, j));
			}
	while(!q.empty()){
		P p = q.front();
		q.pop();
		Int x = p.first;
		Int y = p.second;
		for(Int i=0; i<4; ++i){
			Int nx = x + dx[i];
			Int ny = y + dy[i];
			if(0 <= nx && nx < h && 0 <= ny && ny < w && a[nx][ny] == '.' && v[x][y] - 1 > v[nx][ny]){
				v[nx][ny] = v[x][y] - 1;
				if(v[nx][ny] > 0) q.push(P(nx, ny));
			}
		}
	}
	Int ans = 1e9;
	for(Int i=0; i<h; ++i)
		for(Int j=0; j<w; ++j)
			if(v[i][j] != -1){
				if(i == 0 || j == 0 || i == h-1 || j == w-1) ans = 1;
				else ans = min({ans, 1+(i+k-1)/k, 1+(j+k-1)/k, 1+(h-1-i+k-1)/k, 1+(w-1-j+k-1)/k});
			}
	cout << ans << "\n";
//	for(Int i=0; i<h; ++i)
//		for(Int j=0; j<w; ++j)
//			cout << v[i][j] << (" \n"[j+1==w]);
}