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;
#define ll long long
#define all(V) V.begin(),V.end()
#define FINISH cout<<"No"<<endl;return 0;

bool arr[810][810];
string ma[810];
ll H, W;
ll hi = LLONG_MAX / 2, wi = hi, hx = 0, wx = 0;

int main() {
	ll K, a, b, c;
	cin >> H >> W >> K;
	for (int i = 0;i < H;i++)cin >> ma[i];
	for (int i = 0;i < H;i++) {
		for (int j = 0;j < W;j++) {
			arr[i][j] = 0;
			if (ma[i][j] == 'S') {
				a = i, b = j;
			}
		}
	}
	queue<tuple<ll, ll, ll>> Q;
	Q.push(make_tuple(a, b, K));
	while (!Q.empty()) {
		tie(a, b, c) = Q.front();
		Q.pop();
		if (ma[a][b] == '#') {
			continue;
		}
		if (a == 0 || b == 0 || a == H - 1 || b == W - 1) {
			cout << 1 << endl;
			return 0;
		}
		if (arr[a][b])continue;
		arr[a][b] = 1;
		hi = min(hi, a);
		wi = min(wi, b);
		hx = max(hx, a);
		wx = max(wx, b);
		if (c < 1)continue;
		Q.push(make_tuple(a - 1, b, c - 1));
		Q.push(make_tuple(a + 1, b, c - 1));
		Q.push(make_tuple(a, b - 1, c - 1));
		Q.push(make_tuple(a, b + 1, c - 1));
	}
	hi = min(hi, H - hx - 1);
	wi = min(wi, W - wx - 1);
	c = min(hi, wi);
	cout << (c - 1) / K + 2 << endl;
}