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 <queue>
#include <iostream>
using namespace std;
int main() {
    int h, w, k; cin >> h >> w >> k;
    string s[h]; for (int i = 0; i < h; i++) cin >> s[i];
    int sx, sy;
    for (int x = 0, done = 0; x < h && !done; x++)
        for (int y = 0; y < w && !done; y++)
            if (s[x][y] == 'S') { sx = x; sy = y; done = 1; }
    priority_queue<pair<int, pair<int, int> > > pq;
    pq.push({0, {sx, sy}});
    int dist[h][w];
    for (int i = 0; i < h; i++) fill(dist[i], dist[i]+w, 1<<30);
    dist[sx][sy] = 0;
    int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
    for (;;) {
        int d = -pq.top().first,
            x = pq.top().second.first, y = pq.top().second.second;
        pq.pop();
        if (d > dist[x][y]) continue;
        if (x == 0 || x == h-1 || y == 0 || y == w-1) {
            cout << (d + k - 1) / k << endl;
            return 0;
        }
        for (int i = 0; i < 4; i++) {
            int x1 = x + dx[i], y1 = y + dy[i];
            int d1 = (s[x1][y1] == '#' && d < k ? k+1 : d+1);
            if (dist[x1][y1] <= d1) continue;
            dist[x1][y1] = d1;
            pq.push({-d1, {x1, y1}});
        }
    }
}