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 ll MOD = 1e9 + 7;
const double PI = 3.141592653589793238;
const double EPS = 1e-10;
typedef pair<int, int> P;

string A[800];
int dist[800][800];
int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };

int main() {
	int H, W, K;
	cin >> H >> W >> K;
	int sx, sy;
	for (int i = 0; i < H; i++) {
		cin >> A[i];
		for (int j = 0; j < W; j++) {
			if (A[i][j] == 'S') sx = j, sy = i;
		}
	}
	memset(dist, -1, sizeof(dist));
	dist[sy][sx] = 0;
	queue<P> Q;
	Q.push(P(sx, sy));
	int ans = 1 << 30;
	while (!Q.empty()) {
		P p = Q.front(); Q.pop();
		int x = p.first, y = p.second;
		if (dist[y][x] > K) break;
		int d = min(x, W - 1 - x);
		d = min(d, y);
		d = min(d, H - 1 - y);
		ans = min(ans, 1 + (d + K - 1) / K);
		for (int i = 0; i < 4; i++) {
			int nx = x + dx[i];
			int ny = y + dy[i];
			if (0 > nx || nx >= W || 0 > ny || ny >= H || A[ny][nx] == '#' || dist[ny][nx] != -1) continue;
			dist[ny][nx] = dist[y][x] + 1;
			Q.push(P(nx, ny));
		}
	}
	cout << ans << endl;
}