Problem:
Score : 300 points

Problem Statement

There is a sequence of length N: A_1, A_2, ..., A_N. Initially, this sequence is a permutation of 1, 2, ..., N.

On this sequence, Snuke can perform the following operation:

Choose K consecutive elements in the sequence. Then, replace the value of each chosen element with the minimum value among the chosen elements.

Snuke would like to make all the elements in this sequence equal by repeating the operation above some number of times.
Find the minimum number of operations required.
It can be proved that, Under the constraints of this problem, this objective is always achievable.

Constraints

2 \leq K \leq N \leq 100000

A_1, A_2, ..., A_N is a permutation of 1, 2, ..., N.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N

Output

Print the minimum number of operations required.

Sample Input 1

4 3
2 3 1 4

Sample Output 1

2

One optimal strategy is as follows:

In the first operation, choose the first, second and third elements. The sequence A becomes 1, 1, 1, 4.

In the second operation, choose the second, third and fourth elements. The sequence A becomes 1, 1, 1, 1.

Sample Input 2

3 3
1 2 3

Sample Output 2

1

Sample Input 3

8 3
7 3 1 8 4 6 2 5

Sample Output 3

4

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

//参考：https://qiita.com/DaikiSuyama/items/9dba963ba74396b9b7fd
//インクルード(アルファベット順)
#include <algorithm>     //sort,二分探索,など
#include <bitset>        //固定長bit集合
#include <cmath>         //pow,logなど
#include <complex>       //複素数
#include <deque>         //両端アクセスのキュー
#include <functional>    //sortのgreater
#include <iomanip>       //setprecision(浮動小数点の出力の誤差)
#include <iostream>      //入出力
#include <iterator>      //集合演算(積集合,和集合,差集合など)
#include <map>           //map(辞書)
#include <numeric>       //iota(整数列の生成),gcdとlcm(c++17)
#include <queue>         //キュー
#include <set>           //集合
#include <stack>         //スタック
#include <string>        //文字列
#include <unordered_map> //イテレータあるけど順序保持しないmap
#include <unordered_set> //イテレータあるけど順序保持しないset
#include <utility>       //pair
#include <vector>        //可変長配列
using namespace std;
typedef long long ll;
//マクロ
//forループ関係
//引数は、(ループ内変数,動く範囲)か(ループ内変数,始めの数,終わりの数)、のどちらか
//Dがついてないものはループ変数は1ずつインクリメントされ、Dがついてるものはループ変数は1ずつデクリメントされる
#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define REPD(i, n) for (ll i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = a; i <= (ll)(b); i++)
#define FORD(i, a, b) for (ll i = a; i >= (ll)(b); i--)
//xにはvectorなどのコンテナ
#define ALL(x) (x).begin(), (x).end() //sortなどの引数を省略したい
#define SIZE(x) ((ll)(x).size())      //sizeをsize_tからllに直しておく
//定数
#define INF 1000000000000 //10^12:極めて大きい値,∞
#define MOD 1000000007    //10^9+7:合同式の法
#define MAXR 100000       //10^5:配列の最大のrange(素数列挙などで使用)
//略記
#define PB push_back //vectorヘの挿入
#define MP make_pair //pairのコンストラクタ
#define FIR first    //pairの一つ目の要素
#define SEC second   //pairの二つ目の要素
signed main()
{
    //入力の高速化用のコード
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    ll N, K;
    cin >> N >> K;

    vector<int> A(N);
    REP(i, N)
    {
        cin >> A[i];
    }

    //配列中の最小値のある場所を取得
    vector<int>::iterator iter = min_element(A.begin(), A.end());
    size_t index = distance(A.begin(), iter);

    ll ans = 0;
    //左側を計算
    ll LQ = index / (K - 1);
    ll LR = index % (K - 1);
    ans += (LR == 0 ? LQ : LQ + 1);

    //右側を計算
    ll RQ = (N - index - 1) / (K - 1);
    ll RR = (N - index - 1) % (K - 1);
    ans += (RR == 0 ? RQ : RQ + 1);

    cout << ans << endl;

    return 0;
}