Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define debug(x) cout << #x" = " << x;

#define st first
#define nd second

using namespace std;
using namespace placeholders;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;

const int MAXN = 800 + 10;
const int d[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};

int n, m, K;
char s[MAXN][MAXN];
int dis[MAXN][MAXN];

int main(){
	scanf("%d%d%d", &n, &m, &K);
	int sx, sy;
	for (int i = 1; i <= n; ++i){
		scanf("%s", s[i] + 1);
		for (int j = 1; j <= m; ++j)
			if (s[i][j] == 'S')
				sx = i, sy = j;
	}

	queue<PII> Q;
	Q.emplace(sx, sy);
	dis[sx][sy] = 1;
	while (!Q.empty()){
		int x0 = Q.front().st, y0 = Q.front().nd;
		Q.pop();
		for (int i = 0; i < 4; ++i){
			int x = x0 + d[i][0], y = y0 + d[i][1];
			if (s[x][y] == '.' && !dis[x][y]){
				dis[x][y] = dis[x0][y0] + 1;
				if (dis[x][y] < K + 1)
					Q.emplace(x, y);
			}
		}
	}

	int ans = INT_MAX;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			if (dis[i][j]){
				int t = min(min(i - 1, n - i), min(j - 1, m - j));
				ans = min(ans, (t + K - 1) / K + 1);
			}
	printf("%d\n", ans);
	return 0;
}