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 <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const int N = 805, inf = 1e9 + 7 ;
const int flg[4][2] = {{1, 0}, {- 1, 0}, {0, 1}, {0, - 1}} ;
using namespace std ;

int n, m, k, f[N][N] ;
char a[N][N] ;
struct poi {
	int x, y ;
} opt[N * N] ;

bool check(int x, int y) {
	return (a[x][y] != '#' && x >= 1 && y >= 1 && x <= n && y <= m && f[x][y] == inf) ;
}

int main() {
	scanf("%d%d%d", &n, &m, &k) ;
	rep(i, 1, n) scanf("%s", a[i] + 1) ;
	int x, y ;
	rep(i, 1, n) rep(j, 1, m) if (a[i][j] == 'S') x = i, y = j ;
	int he = 0, ta = 1 ;
	rep(i, 1, n) rep(j, 1, m) f[i][j] = inf ;
	opt[1].x = x, opt[1].y = y, f[x][y] = 0 ;
	for ( ; he != ta; ) {
		++ he ;
		int x = opt[he].x, y = opt[he].y ;
		rep(i, 0, 3) {
			int xx = x + flg[i][0], yy = y + flg[i][1] ;
			if (check(xx, yy)) f[xx][yy] = f[x][y] + 1, ++ ta, opt[ta].x = xx, opt[ta].y = yy ; 
		}
	}
	int ans = inf ;
	rep(i, 1, n) rep(j, 1, m) if (f[i][j] <= k) ans = min(ans, (min(i - 1, min(n - i, min(j - 1, m - j))) + k - 1) / k + 1) ;
	printf("%d\n", ans) ;
	return 0 ;
}