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 <iostream>
#include <cmath>
using namespace std;

const int maxn = 808;
int n, m, k, sx, sy, ans = maxn,  bio[maxn][maxn];
char c[maxn][maxn];
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

void solve(int x, int y){
	bio[x][y] = true;

	if(abs(x - sx) + abs(y - sy) > k){
		return;
	}

	for(int i = 0; i < 4; i++){
		int nx = x + dx[i], ny = y + dy[i];

		if(x >= 1 && x <= n && y >= 1 && y <= m && !bio[nx][ny] && c[nx][ny] == '.'){
			solve(nx, ny);
		}
	}

	int d = min(min(x - 1, n - x), min(y - 1, m - y));
	int tmp = d / k + bool(d % k);

	ans = min(ans, tmp);
}

int main(){
	ios_base::sync_with_stdio(false);

	cin >> n >> m >> k;

	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			cin >> c[i][j];

			if(c[i][j] == 'S'){
				sx = i, sy = j;
			}
		}
	}

	solve(sx, sy);

	cout << ans + 1 << endl;

	return 0;
}
