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 <string>
#include <queue>
#include <utility>

using namespace std;

int main()
{
    int H, W, K;
    cin>>H>>W>>K;
    vector<string> A(H);
    for (string &a: A)
        cin>>a;

    vector<vector<int>> D(H, vector<int>(W, 99999999));
    queue<pair<int,int>> Q;
    for (int y=0; y<H; y++)
    for (int x=0; x<W; x++)
        if (A[y][x] == 'S')
            D[y][x] = 0,
            Q.push(make_pair(x, y));
    int dx[] = {-1, 1, 0, 0};
    int dy[] = {0, 0, -1, 1};
    while (!Q.empty())
    {
        int sx = Q.front().first;
        int sy = Q.front().second;
        Q.pop();
        for (int d=0; d<4; d++)
        {
            int tx = sx + dx[d];
            int ty = sy + dy[d];
            if (0<=tx && tx<W && 0<=ty && ty<H && A[ty][tx]=='.' && D[ty][tx]>D[sy][sx]+1)
                D[ty][tx] = D[sy][sx]+1,
                Q.push(make_pair(tx, ty));
        }
    }

    int ans = 99999999;
    for (int y=0; y<H; y++)
    for (int x=0; x<W; x++)
        if (D[y][x] <= K)
            ans = min(ans, (min(min(y,H-y-1),min(x,W-x-1))+K-1)/K + 1);
    cout<<ans<<endl;
}