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
#include <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007, INF=mod*mod*3LL;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;
struct node{ll i, j, dis;};
int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll h, w, m, sx, sy;
    cin >> h >> w >> m;
    string s[h];

    for(int i=0;i<h;i++){
    	cin >> s[i];
    	for(int j=0;j<w;j++){
    		if(s[i][j] == 'S') sx = j, sy = i;
    	}
    }

    ll dx[4] = {-1, 0, 0, 1}, dy[4] = {0, -1, 1, 0};
    ll ans = 1e9;

    queue<node> q;
    q.push({sy, sx, 0});
    bool used[h][w];
    memset(used,0,sizeof(used));
    used[sy][sx] = true;
    while(!q.empty()){
    	node now = q.front(); q.pop();
    	ans = min({ans, now.i, now.j, w - now.j - 1, h - now.i - 1});
    	if(now.dis == m) continue;
    	for(int k=0;k<4;k++){
    		ll y = now.i + dy[k];
    		ll x = now.j + dx[k];
    		if(x < 0 || y < 0 || x >= w || y >= h || s[y][x] == '#' || used[y][x]) continue;
    		used[y][x] = true;
    		q.push({y, x, now.dis + 1});
    	}
    }
    cout << (ans + m * 2 - 1) / m << endl;
    return 0;
}
