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
# https://beta.atcoder.jp/contests/agc014/submissions/1268295
from collections import deque

dr = (0, 1, 0, -1)
dc = (1, 0, -1, 0)


def solve():
    H, W, K = map(int, input().split())
    a = [[False for _ in range(W)] for _ in range(H)]

    sy, sx = -1, -1
    for r in range(H):
        s = input()
        for c, x in enumerate(s):
            if x == '.':
                a[r][c] = True
            elif x == 'S':
                a[r][c] = True
                sy, sx = r, c

    visited = [[False for _ in range(W)] for _ in range(H)]
    visited[sy][sx] = True

    t = max(H, W)
    q = deque([(sy, sx, 0)])
    while q:
        r, c, k = q.popleft()

        t = min(t, r, H - 1 - r, c, W - 1 - c)
        # print(t)
        if t == 0:
            print(1)
            return

        for d_i in range(4):
            nr, nc = r + dr[d_i], c + dc[d_i]
            if 0 <= nr < H and 0 <= nc < W and a[nr][nc] and not visited[nr][nc] and k + 1 <= K:
                visited[nr][nc] = True
                q.append((nr, nc, k + 1))

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

    print(ans)


if __name__ == '__main__':
    solve()
