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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;

LL vis[805][805] = {0};
char ch[805][805];
LL n,m,k,ans = INF;
LL dx[4] = {0,0,1,-1};
LL dy[4] = {1,-1,0,0};

struct node{
	LL x,y,stp;
}h,t; queue <node> q;
void bfs(LL x,LL y){
	t.x = x; t.y = y; t.stp = 0; q.push(t);
	while(!q.empty()){
		h = q.front(); q.pop();
		if(h.stp > k) continue;
		vis[h.x][h.y] = 1;
		ans = min(ans,min(min(h.x - 1,h.y - 1),min(n - h.x,m - h.y)));
		for(LL i = 0;i < 4;i ++){
			LL tx = h.x + dx[i],ty = h.y + dy[i];
			if(!tx || !ty || tx > n || ty > m) continue;
			if(vis[tx][ty] || ch[tx][ty] == '#') continue;
			vis[tx][ty] = 1;
			t.x = tx; t.y = ty; t.stp = h.stp + 1;
			q.push(t);
		}
	}
}

int main(){
	ios::sync_with_stdio(false);
	LL sx = 0,sy = 0;
	cin >> n >> m >> k;
	for(LL i = 1;i <= n;i ++){
		for(LL j = 1;j <= m;j ++){
			cin >> ch[i][j];
			if(ch[i][j] == 'S'){ sx = i; sy = j; }
		}
	}
	bfs(sx,sy);
//	cout << ans << endl;
	cout << (ans + k - 1) / k + 1 << endl;
	return 0;
}