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<stdio.h>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
typedef pair<int, int>pii;
typedef pair<int, pii>pi3;
int m[1000][1000];
bool flag[1000][1000];
int dx[4] = { 1,0,-1,0 };
int dy[4] = { 0,1,0,-1 };
int main()
{
	int mx, my, k;
	scanf("%d%d%d", &mx, &my, &k);
	int sx, sy;
	for (int i = 0; i < mx; i++)
	{
		for (int j = 0; j < my; j++)
		{
			char z;
			scanf(" %c", &z);
			if (z != '#')m[i][j] = 1;
			if (z == 'S')sx = i, sy = j;
		}
	}
	queue<pi3>que;
	que.push(make_pair(0, make_pair(sx, sy)));
	int mini = 1000000000;
	for (;;)
	{
		if (que.empty())break;
		pi3 z = que.front();
		que.pop();
		int d = z.first, x = z.second.first, y = z.second.second;
		if (flag[x][y])continue;
		flag[x][y] = true;
		mini = min(mini, (x + k - 1) / k);
		mini = min(mini, (y + k - 1) / k);
		mini = min(mini, (mx - x - 1 + k - 1) / k);
		mini = min(mini, (my - y - 1 + k - 1) / k);
		if (d == k)continue;
		for (int i = 0; i < 4; i++)if (m[x + dx[i]][y + dy[i]] == 1)que.push(make_pair(d + 1, make_pair(x + dx[i], y + dy[i])));
	}
	printf("%d\n", mini + 1);
}