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;
using pr = pair<int, int>;

int N, M, K, ans = 1e9;
int d[4] = { 1, -1, 0, 0 }, Sx, Sy;
int dist[808][808];
char A[808][808];
queue<pr> q;

int f(int x, int y) { return min({ x - 1, y - 1, N - x, M - y }); }

int main() {
    scanf("%d %d %d", &N, &M, &K);
    for(int i=1; i<=N; i++) scanf("%s", A[i] + 1);
    for(int i=1; i<=N; i++) for(int j=1; j<=M; dist[i][j] = 1e9, j++) if(A[i][j] == 'S') Sx = i, Sy = j;
    q.push(pr(Sx, Sy)), dist[Sx][Sy] = 0;
    while(!q.empty()) {
        int x = q.front().first, y = q.front().second; q.pop();
        ans = min(ans, f(x, y));
        if(dist[x][y] == K) continue;
        for(int i=0; i<4; i++) {
            int xx = x + d[i], yy = y + d[3-i];
            if(A[xx][yy] == '.' && dist[xx][yy] > dist[x][y] + 1) {
                q.push(pr(xx, yy));
                dist[xx][yy] = dist[x][y] + 1;
            }
        }
    }
    printf("%d\n", ans / K + !!(ans % K) + 1);
}