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>
using namespace std;
typedef long long ll;

const int INF = 1 << 28;
int d[800][800];
string b[800];
int dx[4] = { 1, 0, -1, 0 }, dy[4] = { 0, 1, 0, -1 };

struct S {
	int x, y;
};
int H, W, K;

bool goal(int x, int y) {
	return x == 0 || x == W - 1 || y == 0 || y == H - 1;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> H >> W >> K;

	int sx = -1, sy = -1;
	for(int i = 0; i < H; i++) {
		cin >> b[i];
		for(int j = 0; j < W; j++) {
			if(b[i][j] == 'S') sx = j, sy = i;
		}
	}

	fill((int*)begin(d), (int*)end(d), INF);
	d[sy][sx] = 0;

	int ans = INF;
	queue<S> q;
	q.push({ sx, sy });
	while(q.size()) {
		S s = q.front();
		q.pop();
		int mind = min({ s.x, s.y, W - 1 - s.x, H - 1 - s.y });
		ans = min(ans, (mind + K - 1) / K + 1);

		if(d[s.y][s.x] >= K) continue;
		for(int k = 0; k < 4; k++) {
			int nx = s.x + dx[k], ny = s.y + dy[k];
			if(b[ny][nx] == '#') continue;
			if(goal(nx, ny)) {
				cout << 1 << endl;
				return 0;
			}
			if(d[ny][nx] > d[s.y][s.x] + 1) {
				d[ny][nx] = d[s.y][s.x] + 1;
				q.push({ nx, ny });
			}
		}
	}

	cout << ans << endl;
}