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
import math
import os
import sys
from collections import deque

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7

H, W, K = list(map(int, sys.stdin.readline().split()))
A = [sys.stdin.readline().rstrip() for _ in range(H)]

sh = sw = -1
for h, r in enumerate(A):
    if 'S' in r:
        sh = h
        sw = r.index('S')


def step(h, w):
    return math.ceil(min(h, w, H - h - 1, W - w - 1) / K)


def ok(h, w):
    return 0 <= h < H and 0 <= w < W


dh = [0, 1, 0, -1]
dw = [1, 0, -1, 0]

# 最初の魔法でどこまで端にいけるか
que = deque([(sh, sw, K)])
seen = [[0] * W for _ in range(H)]
seen[sh][sw] = True
min_step = step(sh, sw)
while que:
    sh, sw, k = que.popleft()
    for hh, ww in zip(dh, dw):
        h = sh + hh
        w = sw + ww
        if not ok(h, w) or A[h][w] == '#' or seen[h][w]:
            continue
        if k > 0:
            min_step = min(min_step, step(h, w))
            seen[h][w] = True
            que.append((h, w, k - 1))
print(min_step + 1)
