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 h, w, k;
const int INF = 1e9;
const int N = 850;
string A[N]; int vis[N][N];
int ans = INF; queue <int> Q;
int check(int i, int j)
{
    return ans = min(ans, (min(min(i, h - 1 - i), min(j, w - 1 - j)) + k - 1) / k + 1);
}
int main()
{
    cin >> h >> w >> k;
    for (int i = 0; i < h; ++ i)
        cin >> A[i];
    for (int i = 0; i < h; ++ i)
        for (int j = 0; j < w; ++ j)
            if (A[i][j] == 'S')
                Q.push(i), Q.push(j), Q.push(k),
                vis[i][j] = 1;
    while (!Q.empty())
    {
        int i = Q.front(); Q.pop();
        int j = Q.front(); Q.pop();
        int k = Q.front(); Q.pop();
        if (check(i, j) == 1) break;
        if (k == 0) continue;
        for (int di = -1; di <= 1; ++ di)
            for (int dj = -1; dj <= 1; ++ dj)
                if (abs(di) + abs(dj) == 1) if (A[i + di][j + dj] == '.' && !vis[i + di][j + dj])
                    Q.push(di + i), Q.push(dj + j), Q.push(k - 1),
                    vis[i + di][j + dj] = 1;
    }

    cout << ans << endl;
}