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"
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;

using ll = long long;

int main() {
    int H, W, K, si, sj;
    cin >> H >> W >> K;
    vector<string> A(H);
    rep(i, H) {
        cin >> A[i];
        rep(j, W) {
            if (A[i][j] == 'S') {
                si = i;
                sj = j;
            }
        }
    }
    vector<vector<int>> d(H, vector<int>(W, 1e9));
    d[si][sj] = 0;
    queue<vector<int>> que;
    que.push({si, sj});
    int m = 1e9;
    while (!que.empty()) {
        int i = que.front()[0], j = que.front()[1];
        que.pop();
        m = min(m, i);
        m = min(m, H - 1 - i);
        m = min(m, j);
        m = min(m, W - 1 - j);
        if (d[i][j] == K) continue;
        rep(k, 4) {
            vector<int> di = {1, 0, -1, 0};
            vector<int> dj = {0, 1, 0, -1};
            int u = i + di[k];
            int v = j + dj[k];
            if (u < 0 || u >= H || v < 0 || v >= W) continue;
            if (d[u][v] < 1e9) continue;
            if (A[u][v] == '#') continue;
            d[u][v] = d[i][j] + 1;
            que.push({u, v});
        }
    }
    cout << (m + K - 1) / K + 1 << endl;
}