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 CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;

int x, y, N, M, K, d[809][809];
char sir[809][809];
queue < pair < int, int > > cc;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d %d %d\n", &N, &M, &K);
for (int i=1; i<=N; i++)
{
    scanf ("%s", sir[i] + 1);
    for (int j=1; j<=M; j++)
    {
        if (sir[i][j] == 'S')
            x = i, y = j, sir[i][j] = '.';
        d[i][j] = -1;
    }
}
d[x][y] = 0;
cc.push ({x, y});
while (!cc.empty ())
{
    auto curr = cc.front ();
    cc.pop ();
    for (int k=0; k<4; k++)
    {
        int x = curr.first + dx[k], y = curr.second + dy[k];
        if (sir[x][y] == '.' && d[x][y] == -1)
            d[x][y] = d[curr.first][curr.second] + 1,
            cc.push ({x, y});
    }
}
int ans = 2 * N + 2 * M + 2 * K;
for (int i=1; i<=N; i++)
    for (int j=1; j<=M; j++)
        if (d[i][j] != -1 && d[i][j] <= K)
        {
            int fromHere = min (min (i - 1, N - i), min (j - 1, M - j)), t = fromHere / K + (fromHere % K != 0);
            if (t + 1 < ans)
                ans = t + 1;
        }
printf ("%d\n", ans);
return 0;
}