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 <queue>
#define inf 1000000000

using namespace std;

int H, W, K;
char map[805][805];
int sx, sy;
int dist[805][805];
const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};

int bfs()
{
	for(int x = 0; x < W; x++){
		for(int y = 0; y < H; y++){
			dist[x][y] = inf;
		}
	}
	dist[sx][sy] = 0;
	
	int ret = inf, d;
	queue<int> Q;
	Q.push(sx*1000+sy);
	int x, y, nx, ny;
	while(Q.size()){
		x = Q.front() / 1000;
		y = Q.front() % 1000;
		Q.pop();
		d = min(x, y);
		d = min(d, W-x-1);
		d = min(d, H-y-1);
		ret = min(ret, d);
		if(dist[x][y] == K) continue;
		for(int d = 0; d < 4; d++){
			nx = x + dx[d], ny = y + dy[d];
			if(nx < 0 || nx >= W || ny < 0 || ny >= H) continue;
			if(map[nx][ny] == '#') continue;
			if(dist[nx][ny] != inf) continue;
			Q.push(nx*1000+ny);
			dist[nx][ny] = dist[x][y] + 1;
		}
	}
	return ret;
}

int main(void)
{
	cin >> H >> W >> K;
	for(int y = 0; y < H; y++){
		for(int x = 0; x < W; x++){
			cin >> map[x][y];
			if(map[x][y] == 'S'){
				sx = x, sy = y;
				map[x][y] = '.';
			}
		}
	}
	
	int ans = bfs();
	cout << (ans+K-1) / K + 1<< endl;
	return 0;
}