Gym

Doomsday comes in t units of time. In anticipation of such a significant eventn people prepared m vaults in which, as they think, it will be possible to survive. But each vault can accommodate onlyk 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 andt (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 and10⁹, inclusively.

Output

Output one integer — the maximal number of people who can hide in vaults and think they will survive.

Example

Input

2 2 1 545 5540 60

Output

2

Input

2 2 1 545 5440 60

Output

1

Input

2 2 2 545 3540 60

Output

2

Input

3 3 1 540 45 4545 50 50

Output

3

思路：贪心问题，把棺材作为研究对象，从左到右遍历一遍即可；

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int peo[200005],res[200005];int main(){int n,m,k,t;while(scanf("%d %d %d %d",&n,&m,&k,&t)!=EOF){for(int i=0;i<n;i++)scanf("%d",&peo[i]);for(int i=0;i<m;i++)scanf("%d",&res[i]);sort(peo,peo+n);sort(res,res+m);int now=0,cnt=0,ans=0;for(int i=0;i<m;i++){cnt=k;int j=now;for(;peo[j]<=res[i]+t && j<n && cnt;j++){if(peo[j]>=res[i]-t){ans++;cnt--;}}now=j;if(j==n) break;}printf("%d\n",ans);}return 0;}