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 main(){
    int H, W, K;
    cin >> H >> W >> K;
    string S[800];
    for(int i=0; i<H; i++) cin >> S[i];
    int dist[800][800];
    queue<pair<int, int>> que;
    for(int i=0; i<H; i++) for(int j=0; j<W; j++){
        dist[i][j] = 1e9;
        if(S[i][j] == 'S'){
            que.push({i, j});
            dist[i][j] = 0;
        }
    }

    int dx[] = {0, 0, 1, -1};
    int dy[] = {1, -1, 0, 0};
    auto inside = [&](int i, int j){ return i>=0 && i<H && j>=0 && j<W; };

    while(que.size()){
        auto p = que.front(); que.pop();
        int i = p.first, j = p.second;
        for(int k=0; k<4; k++){
            int x = i+dx[k];
            int y = j+dy[k];
            if(inside(x, y) && S[x][y] != '#' && dist[x][y] == 1e9){
                dist[x][y] = dist[i][j] + 1;
                que.push({x, y});
            }
        }
    }

    int ans = 1e9;
    for(int i=0; i<H; i++) for(int j=0; j<W; j++) if(dist[i][j] <= K){
        int d = min({i, H-1-i, j, W-1-j});
        ans = min(ans, (d+K-1)/K + 1);
    }
    cout << ans << endl;
}