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<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

typedef pair<int,int> pii;
const int MAXN=805;
const int INF=0x3f3f3f3f;
#define x first
#define y second

int n,m,K,ans=INF;
int vis[MAXN][MAXN];
char s[MAXN][MAXN];
pii S;
int df[4][2]={{0,-1},{0,1},{-1,0},{1,0}};

int f(int x,int y){
    int len=min(min(x-1,y-1),min(n-x,m-y));
    return len/K+((len%K)?1:0);
}

queue <pii> q;
void bfs(){
    memset(vis,0,sizeof vis);
    q.push(S);
    vis[S.x][S.y]=1;
    while(!q.empty()){
        pii now=q.front(); q.pop();
        ans=min(ans,f(now.x,now.y));
        for(int i=0;i<4;i++){
            int nx=now.x+df[i][0];
            int ny=now.y+df[i][1];
            if(nx<1||ny<1||nx>n||ny>m||vis[nx][ny]||abs(nx-S.x)+abs(ny-S.y)>K||s[nx][ny]=='#') continue;
            vis[nx][ny]=1;
            q.push(make_pair(nx,ny));
        }
    }
}

int main(){

    scanf("%d%d%d",&n,&m,&K);
    for(int i=1;i<=n;i++){
        scanf("%s",s[i]+1);
        for(int j=1;j<=m;j++)
            if(s[i][j]=='S') S.x=i,S.y=j;
    }
    if(S.x==1||S.x==n||S.y==1||S.y==m) return puts("0"),0;
    bfs();
    printf("%d",ans+1);

    fclose(stdin);
    fclose(stdout);
}