CF-1055E：Segments on the Line （二分&背包&DP优化）（nice problem）

You are a given a list of integers a 1 ,a 2 ,…,a n  a1,a2,…,an and s s of its segments [l j ;r j ] [lj;rj] (where 1≤l j ≤r j ≤n 1≤lj≤rj≤n ).

You need to select exactly m m segments in such a way that the k k -th order statistic of the multiset of a i  ai , where i i is contained in at least one segment, is the smallest possible. If it's impossible to select a set of m m segments in such a way that the multiset contains at least k k elements, print -1.

The k k -th order statistic of a multiset is the value of the k k -th element after sorting the multiset in non-descending order.

Input

The first line contains four integers n n , s s , m m and k k (1≤m≤s≤1500 1≤m≤s≤1500 , 1≤k≤n≤1500 1≤k≤n≤1500 ) — the size of the list, the number of segments, the number of segments to choose and the statistic number.

The second line contains n n integers a i  ai (1≤a i ≤10 9  1≤ai≤109 ) — the values of the numbers in the list.

Each of the next s s lines contains two integers l i  li and r i  ri (1≤l i ≤r i ≤n 1≤li≤ri≤n ) — the endpoints of the segments.

It is possible that some segments coincide.

Output

Print exactly one integer — the smallest possible k k -th order statistic, or -1 if it's impossible to choose segments in a way that the multiset contains at least k k elements.

Examples

Input

4 3 2 2
3 1 3 2
1 2
2 3
4 4

Output

2

Input

5 2 1 1
1 2 3 4 5
2 4
1 5

Output

1

Input

5 3 3 5
5 5 2 1 1
1 2
2 3
3 4

Output

-1

题意：给定给N个点，以及M个线段，让你选择S个线段，使得至少被一个线段覆盖的点排序后，第K大最小，没有则输出-1。

思路：求第K大最小，显然需要二分，每次验证看当前的mid是否有大于等于K个数小于mid。验证我们用dp来验证，复杂度是O(NMS*lgN);

需要优化掉一个。这里用背包把M优化掉了，我们找到每个点的Next，Next代表包含这个点的最右端。就不难得到dp方程，这个时候M已经没用了。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=1510;
struct in{ int L,R;}s[maxn];
int a[maxn],b[maxn],N,S,M,K,sum[maxn];
int dp[maxn][maxn],Next[maxn];
bool check(int Mid) //M个选最多S个的第K大
{
    rep(i,1,N) sum[i]=sum[i-1]+(a[i]<=Mid);
    rep(i,0,S) rep(j,0,N) dp[i][j]=0;
    rep(i,1,S){
        rep(j,1,N) dp[i][j]=max(dp[i][j],dp[i-1][j]); //不选j位置。
        rep(j,1,N) if(Next[j]) dp[i][Next[j]]=max(dp[i][Next[j]],dp[i-1][j-1]+sum[Next[j]]-sum[j-1]); //选j
        rep(j,1,N) dp[i][j]=max(dp[i][j],dp[i][j-1]);
    }
    return dp[S][N]>=K;
}
int main()
{
    scanf("%d%d%d%d",&N,&M,&S,&K);
    rep(i,1,N) scanf("%d",&a[i]),b[i]=a[i];
    rep(i,1,M) scanf("%d%d",&s[i].L,&s[i].R);
    rep(i,1,M) rep(j,s[i].L,s[i].R) Next[j]=max(Next[j],s[i].R);
    sort(b+1,b+N+1); int L=1,R=N,Mid,ans=-1;
    while(L<=R){
        Mid=(L+R)>>1;
        if(check(b[Mid])) ans=b[Mid],R=Mid-1;
        else L=Mid+1;
    }
    printf("%d
",ans);
    return 0;
}