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
def main():
    Dy = [1,0,-1,0]
    Dx = [0,-1,0,1]
    
    H, W, K = map(int, input().split())
    A = [input() for _ in range(H)]
    Closed = [[False]*W for _ in range(H)]
    for y, a in enumerate(A):
        for x, aa in enumerate(a):
            if aa=="S":
                ys, xs = y, x
    q = [(ys, xs)]
    Closed[ys][xs] = True
    reachable = [(ys, xs)]
    for _ in range(K):
        q_new = []
        for y_old, x_old in q:
            for dy, dx in zip(Dy, Dx):
                y, x = y_old + dy, x_old + dx
                if not (0<=y<H and 0<=x<W):
                    continue
                if not Closed[y][x] and A[y][x]==".":
                    Closed[y][x] = True
                    reachable.append((y, x))
                    q_new.append((y, x))
        q = q_new
    mi = float("inf")
    for y, x in reachable:
        mi = min(mi, x, y, W-1-x, H-1-y)
    print((mi-1)//K+2)

main()
