D. One-Dimensional Battle Ships

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

Examples

Input

Copy

11 3 3
5
4 8 6 1 11

Output

Copy

Input

Copy

5 1 3
2
1 5

Output

Copy

-1

Input

Copy

5 1 3
1
3

Output

Copy

题意：

给你一艘船的长度和船的数量，每一艘船至少格一个空格，然后再给你m个查询，问在第几个查询出错了。

思路：

很明显二分，至于怎样二分，其实解题无非迭代或递归，一般二分都是迭代解题，故假设如果这n个查询是有序的，这就很好处理了，故我们先排一个序，再迭代就行。

code:

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int n, k, a,m;
struct node {int x;int id;
}arr[200100];
bool cmp(node a, node b)
{return a.x < b.x;
}
void input()
{cin >> n >> k >> a;cin >> m;for (int i = 1; i <= m; i++){cin >> arr[i].x;arr[i].id = i;}sort(arr + 1, arr + 1 + m, cmp);
}
bool pang(int l)
{int can = 0, yu = 0;for (int i = 1; i <= m; i++){if (arr[i].id <= l){can += (arr[i].x - yu) / (a+1);yu = arr[i].x;}}can += (n - yu+1) / (a + 1);if (can >= k)return 1;elsereturn 0;
}
int solve()
{int l = 1, r = m,ans=0;while (l <= r){int mid = (l + r) / 2;if (pang(mid)){ans = mid;l = mid+1;}else{r = mid - 1;}}return ans;
}
int main()
{input();int ans=solve();if (ans == m)cout << -1 << endl;elsecout << ans + 1 << endl;return 0;
}

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