D - 船之战

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, ..., x_m, where x_i 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 1 to m in the order the were made. If the sought move doesn't exist, then print "-1".

Example

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
题目意思:
有一条1 * n 长的海,k艘1 * a的船,船都在海上,其船不能重叠和靠在一起;
然后会有m发炮弹打在x1,x2....xm处
问第几发炮弹一定会打到船
解法:
每有一发炮弹,就会把海进行分割,计算每个小的海域可以有几艘船,如果之和<k说明一定打到了

#include <iostream>
#include <string.h>
#include <set>
#include <stack>
using namespace std;

const int MAX = 5*100000 + 2000;

int main()
{
    int ti =-1;
    set<int>s;
    int L,n,l;
    int N;
    cin>>L>>n>>l>>N;
    s.insert(0);
    s.insert(L+1);
    int sum = (L+1)/(l+1);
    for(int i = 0;i < N;i++)
    {
        int temp;
        cin>>temp;
        set<int>::iterator it;
        if(!s.empty())
        {
           　　　　int temp1,temp2;
            　　　　it = s.lower_bound(temp);
           　　　　 temp1 = *it ;
          　　　　  temp2 = *(--it);
          　　　　  sum = sum - (temp1 -temp2)/(l+1) + (temp - temp2)/(l+1) +(temp1 - temp)/(l+1); //这个公式很重要
       //     cout<<"temp1 == temp2 =sum == "<<temp1<<temp2<<sum<<endl;
        }
        s.insert(temp);

       　 if(sum < n)
        {
           　　 ti = i+1;
           　　 break;
        }
    }

    cout<<ti<<endl;

    return 0;
}