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<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef pair<int,int> pii;

#define inRange(x,a,b) (a <= x && x < b)
int di[8] = {0,0,1,-1,1,1,-1,-1};
int dj[8] = {1,-1,0,0,1,-1,1,-1};

int dp[800][800];

int main(){
    int h, w, k;
    cin >> h >> w >> k;
    char mat[h][w];
    int si, sj;
    for(int i = 0; i < h; i++){
        for(int j = 0; j < w; j++){
            cin >> mat[i][j];
            if(mat[i][j] == 'S')    si = i, sj = j;
        }
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[si][sj] = 0;
    queue<pii> q;
    q.push({si,sj});
    int ans = 1<<30;
    while(!q.empty()){
        pii p = q.front();  q.pop();
        int i = p.first, j = p.second;
        ans = min(ans, (min({i, h-1-i, j, w-1-j})+k-1)/k);
        if(dp[i][j] == k)   continue;
        for(int k = 0; k < 4; k++){
            int ni = i + di[k], nj = j + dj[k];
            if(!inRange(ni,0,h) || !inRange(nj,0,w) || mat[ni][nj] == '#' || dp[ni][nj] <= dp[i][j]+1)   continue;
            dp[ni][nj] = dp[i][j]+1;
            q.push({ni,nj});
        }
    }
    cout << ans+1 << endl;
}