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;
typedef pair<int,int> ii;
int dx[] = {1,0,-1,0}, dy[] = {0,1,0,-1};
int main(){
    int h,w,k;
    scanf("%d%d%d",&h,&w,&k);
    char a[h][w];
    int sx, sy;
    for (int i = 0; i < h; i++){
        for (int j = 0; j < w; j++){
            scanf(" %c",&a[i][j]);
            if (a[i][j] == 'S'){
                sx = i, sy = j;
            }
        }
    }
    queue<ii> q;
    int dist[h][w];
    memset(dist,-1,sizeof(dist));
    q.push(ii(sx,sy));
    dist[sx][sy] = 0;
    int ans = h*w;
    while (q.size()){
        ii cur = q.front(); q.pop();
        int x = cur.first, y = cur.second;
        ans = min(ans,min(min(x,h-x-1),min(y,w-y-1)));
        if (dist[x][y] == k) continue;
        for (int k = 0; k < 4; k++){
            int nx = x+dx[k], ny = y+dy[k];
            if (nx >= 0 && nx < h && ny >=0 && ny < w && a[nx][ny] == '.' && dist[nx][ny] == -1){
                dist[nx][ny] = dist[x][y]+1;
                q.push(ii(nx,ny));
            }
        }
    }
    //printf("%d ",ans);
    printf("%d",1+(ans-1)/k+(ans!=0));
}