Codeforces 567D：One-Dimensional Battle Ships（二分）

time limit per test : 1 second

memory limit per test : 256 megabytes

input : standard input

output : standard output

Problem Description

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 \(a (1 ≤ n, k, a ≤ 2·10^5)\) — 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 \(x_1, x_2, ..., x_m\), 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 "\(-1\)".

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

The meaning of problems

A length \ (1 \ times n \) in the region \ (K \) th ship, the length of each ship are \ (A \) , Bob shot \ (m \) different positions for each shots, Alice will tell whether Bob hit.

But because Alice would lie, so when Bob hit, Alice would say no hit

Bob asks Alice found lying earliest when the first few times shooting

Thinking

Because with the increase in the number of shooting, Bob found that the greater the likelihood Alice lying, so you can use to solve the dichotomy

The \ ([1, mid] \ ) Interval sort, then calculate these positions if all does not hit, how many ships can be put down, if more than \ (k \) months, then the current \ (mid \) is feasible , the first part of the query, or query the second half

Code

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
int x[maxn];
int b[maxn];
int n,k,a;
bool check(int mid)
{
    for(int i=1;i<=mid;i++)
        b[i]=x[i];
    sort(b+1,b+1+mid);
    int cnt=0;
    for(int i=1;i<=mid;i++)
        cnt+=(b[i]-b[i-1])/(a+1);   // b[i]和b[i-1]之间可以放多少战舰
    // b[mid]->最后一个位置可以放多少战舰
    cnt+=(n-b[mid]+1)/(a+1);
    if(cnt>=k)
        return true;
    return false;
}
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("/home/wzy/in", "r", stdin);
        freopen("/home/wzy/out", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>k>>a;
    int m;
    cin>>m;
    for(int i=1;i<=m;i++)
    {
        cin>>x[i];
    }
    int l=1,r=m,ans=inf;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(check(mid))
            l=mid+1;
        else
            r=mid-1,ans=min(ans,mid);
    }
    if(ans==inf)
        cout<<-1<<endl;
    else
        cout<<ans<<endl;
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}