Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3 3 3
#.#
#S.
###

Sample Output 1

1

Takahashi can move to room (1,2) in one cast.

Sample Input 2

3 3 3
###
#S#
###

Sample Output 2

2

Takahashi cannot move in the first cast, but can unlock room (1,2).
Then, he can move to room (1,2) in the next cast, achieving the objective in two casts.

Sample Input 3

7 7 2
#######
#######
##...##
###S###
##.#.##
###.###
#######

Sample Output 3

2

Solution in PyPy3 (7.3.0) (judge verdict: Accepted)

class Grid:
    def __init__(self, H, W, grid, road_wall=(".", "#")):
        self.road, self.wall = road_wall
        self.H = H
        self.W = W
        self.grid = grid
        self.parent = [[0] * self.W for _ in range(self.H)]
        self.move = [(1, 0), (-1, 0), (0, 1), (0, -1)]     # 動ける場所の候補

    def bfs(self, start=(0,0), goal=-3, time=0, save=False):
        """
        :param start: スタート地点
        :param goal: ゴール地点
        :param save: True = 前回の探索結果を保持する
        :return: （ループがあっても）最短距離。存在しなければ -1
        """
        if save:
            parent = self.parent
        else:
            parent = [[0] * self.W for _ in range(self.H)]
        p, t = start, time
        parent[p[0]][p[1]] = 1
        next_set = deque([(p, t)])
        while next_set:
            p, t = next_set.popleft()
            h, w = p
            for dh, dw in self.move:
                q = (h + dh, w + dw)
                if q[0] < 0 or q[0] >= self.H or q[1] < 0 or q[1] >= self.W:
                    continue
                if self.grid[q[0]][q[1]] == self.wall:         # 壁があったら進まない
                    continue
                if parent[q[0]][q[1]] != 0:
                    continue
                if t >= K:
                    continue
                #### (debug code) ######
                # self.debug(start, goal, q, t)
                ########################
                if q == goal:
                    return t + 1
                parent[q[0]][q[1]] = 1
                next_set.append((q, t + 1))
        return parent

    def debug(self, start, goal, p, t):
        player = p
        debug_grid = list(list(self.grid[h]) for h in range(self.H))
        if start != (None, None):
            debug_grid[start[0]][start[1]] = "S"
        if goal != -3:
            debug_grid[goal[0]][goal[1]] = "G"
        debug_grid[player[0]][player[1]] = "P"
        print("~~~~~~~~~ t = " + str(t + 1) + "  ~~~~~~~~")
        for debug_h in range(self.H):
            print("".join(str(debug_grid[debug_h][debug_w]) for debug_w in range(self.W)))


###########################################################

from collections import deque
import sys
from math import ceil
import marshal
input = sys.stdin.readline



road, wall = ".", "#"                         # (進行可能を意味する記号, 進行不可を意味する記号)

H, W, K = map(int,input().split())                # 左上は(h,w)=(0,0)、右下は(h,w)=(H-1,W-1)

grid = []
for h in range(H):
    grid.append(input().rstrip())
    for w, s in enumerate(grid[-1]):
        if s == "S":
            start = (h,w)
g = Grid(H, W, grid, (road, wall))
visit = g.bfs(start=start)
res = float("inf")
for h in range(H):
    for w in range(W):
        if visit[h][w]:
            res = min(ceil(h/K)+1,ceil(w/K)+1,ceil((H-1-h)/K)+1,ceil((W-1-w)/K)+1,res)
print(res)