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 <iostream>
#include <queue>

using namespace std;
typedef pair<int,int> pii;


int H,W,K;
char grid[805][805];
int dist[805][805];
int x0,y0;

int main() {
	cin >> H >> W >> K;
	for (int i=0; i<H; i++) {
		for (int j=0; j<W; j++) {
			cin >> grid[i][j];
			if (grid[i][j] == 'S') {
				x0 = i;
				y0 = j;
			}
			dist[i][j] = -1;
		}
	}

	int mind = H*W;
	queue<pii> q;
	q.push(pii(x0,y0));
	while (q.size() > 0) {
		pii cur = q.front();
		q.pop();
		int x=cur.first, y=cur.second;
		if (x < 0 || x >= H || y < 0 || y >= W) continue;
		if (dist[x][y] >= 0 || grid[x][y] == '#') continue;
		dist[x][y] = min(min(x,y),min(H-x-1,W-y-1));
		mind = min(mind,dist[x][y]);
		q.push(pii(x-1,y));
		q.push(pii(x+1,y));
		q.push(pii(x,y-1));
		q.push(pii(x,y+1));
	}

	//cout << mind << endl;
	
	if ((mind+K-1)/K < (dist[x0][y0]+K-1)/K)
		cout << (dist[x0][y0]+K-1)/K << "\n";
	else
		cout << (dist[x0][y0]+K-1)/K + 1 << "\n";
}