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 PYTHON3):

Solution:
import sys
from collections import deque

mod = 10**9 + 7
inf = 1<<30

def solve():
    H, W, K = map(int, sys.stdin.readline().split())

    A = [[False]*W for i in range(H)]
    si = sj = -1

    for i in range(H):
        ai = sys.stdin.readline().rstrip()

        for j in range(W):
            if ai[j] == 'S':
                si = i
                sj = j
                A[i][j] = True
            elif ai[j] == '.':
                A[i][j] = True

    ans = bfs(H, W, K, A, si, sj)

    print(ans)

def bfs(H, W, K, A, si, sj):
    dx = (1, 0, -1, 0,)
    dy = (0, 1, 0, -1,)

    min_step = inf

    visited = [[False]*W for i in range(H)]
    visited[si][sj] = True
    nxts = deque([(si, sj, 0)])

    while nxts:
        r, c, d = nxts.popleft()

        min_step = min(min_step, r, H - 1 - r, c, W - 1 - c)

        if min_step == 0:
            return 1

        if d + 1 > K:
            continue

        for k in range(len(dx)):
            nr = r + dy[k]
            nc = c + dx[k]

            if 0 <= nr < H and 0 < nc < W:
                if A[nr][nc] and not visited[nr][nc]:
                    visited[nr][nc] = True
                    nxts.append((nr, nc, d + 1))

    ans = (min_step + K - 1) // K + 1

    return ans

if __name__ == '__main__':
    solve()