Doomsday comes in t units of time. In anticipation of such a significant event n people prepared m vaults in which, as they think, it will be possible to survive. But each vault can accommodate only k people and each person can pass only one unit of distance per one unit of time. Fortunately, all people and vaults are now on the straight line, so there is no confusion and calculations should be simple.
You are given the positions of the people and the vaults on the line. You are to find the maximal number of people who can hide in vaults and think they will survive.
Input
The first line contains four integers n, m, k and t (1 ≤ n, m, k ≤ 200000, 1 ≤ t ≤ 109) separated by spaces — the number of people, the number of vaults, the capacity of one vault and the time left to the Doomsday.
The second line contains n integers separated by spaces — the coordinates of the people on the line.
The third line contains m integers separated by spaces — the coordinates of the vaults on the line.
All the coordinates are between - 109 and 109, inclusively.
Output
Output one integer — the maximal number of people who can hide in vaults and think they will survive.
Example
Input
2 2 1 5
45 55
40 60
Output
2
Input
2 2 1 5
45 54
40 60
Output
1
Input
2 2 2 5
45 35
40 60
Output
2
Input
3 3 1 5
40 45 45
45 50 50
Output
3
题意：世界末日要来了，有n个人，m个地下室，每个地下室可以有k个人，共有t个时间单位，每个单位可以走一个格，都在一条直线上，问最多有多少能活下来。
首先对人和地下室先从小到大排序，然后考虑每个地下室，从1-m个，然后考虑人的情况，当人在范围里面总和就加1
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 200010
using namespace std;
int num[N];
int val[N]; //地下室
int main()
{
int n,m,k,t;
scanf("%d%d%d%d",&n,&m,&k,&t);
for(int i=0;i<n;i++)
scanf("%d",&num[i]);
for(int i=0;i<m;i++)
scanf("%d",&val[i]);
sort(num,num+n);
sort(val,val+m);
int sum=0;
int j=0;
for(int i=0;i<m;i++)
{
int flag=k;
for(;num[j]<=val[i]+t && j<n && flag;j++)
{
if(num[j]>=val[i]-t)
{
flag--;
sum++;
}
}
if(j==n)
break;
}
printf("%d\n",sum);
return 0;
}