CodeForces - 567D One-Dimensional Battle Ships （二分）

題目鏈接：http://codeforces.com/problemset/problem/567/D點擊打開鏈接

D. One-Dimensional Battle Ships

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Alice and Bob love playing one-dimensional battle ships. They play on the field in the form of a line consisting of n square cells (that is, on a 1 × n table).

At the beginning of the game Alice puts k ships on the field without telling their positions to Bob. Each ship looks as a 1 × a rectangle (that is, it occupies a sequence of a consecutive squares of the field). The ships cannot intersect and even touch each other.

After that Bob makes a sequence of "shots". He names cells of the field and Alice either says that the cell is empty ("miss"), or that the cell belongs to some ship ("hit").

But here's the problem! Alice like to cheat. May be that is why she responds to each Bob's move with a "miss".

Help Bob catch Alice cheating — find Bob's first move, such that after it you can be sure that Alice cheated.

Input

The first line of the input contains three integers: n, k and a (1 ≤ n, k, a ≤ 2·105) — the size of the field, the number of the ships and the size of each ship. It is guaranteed that the n, k and a are such that you can put k ships of size a on the field, so that no two ships intersect or touch each other.

The second line contains integer m (1 ≤ m ≤ n) — the number of Bob's moves.

The third line contains m distinct integers x1, x2, ..., xm, where xi is the number of the cell where Bob made the i-th shot. The cells are numbered from left to right from 1 to n.

Output

Print a single integer — the number of such Bob's first move, after which you can be sure that Alice lied. Bob's moves are numbered from 1to m in the order the were made. If the sought move doesn't exist, then print "-1".

Examples

input

11 3 3
5
4 8 6 1 11

output

3

input

5 1 3
2
1 5

output

-1

input

5 1 3
1
3

output

1

注意兩個船之間至少間隔一個空隙！！

沒注意看然後用巨麻煩的模擬搞了一個多小時結果不可維護，，

這道題用set記錄點對點處理而不用記錄區間

然後二分找點所在的區間即可

#include <bits/stdc++.h>
using namespace std;
vector<int >re;
set<int > s;
set<int >::iterator up,down;
int main()
{
    int n,k,a,m;
    cin >> n >>k>> a>> m;
    for(int i=0;i<m;i++)
    {
        int mid;
        scanf("%d",&mid);
        re.push_back(mid);
    }
    s.insert(0);
    s.insert(n+1);
    int cnt=(n+1)/(a+1);
    int ans=-1;
    for(int i=0;i<m;i++)
    {
        int mid=re[i];
        up=s.upper_bound(mid);
        down=(--up);
        up++;
        cnt=cnt-(*up-*down)/(a+1)+(mid-*down)/(a+1)+(*up-mid)/(a+1);
        s.insert(mid);
        if(cnt<k)
        {
            ans=i+1;
            break;
        }
    }
    cout <<ans <<endl;
}