codeforces_738C_二分

C. Road to Cinema

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya is currently at a car rental service, and he wants to reach cinema. The film he has bought a ticket for starts in tminutes. There is a straight road of length s from the service to the cinema. Let's introduce a coordinate system so that the car rental service is at the point 0, and the cinema is at the point s.

There are k gas stations along the road, and at each of them you can fill a car with any amount of fuel for free! Consider that this operation doesn't take any time, i.e. is carried out instantly.

There are n cars in the rental service, i-th of them is characterized with two integers ci and vi — the price of this car rent and the capacity of its fuel tank in liters. It's not allowed to fuel a car with more fuel than its tank capacity vi. All cars are completely fueled at the car rental service.

Each of the cars can be driven in one of two speed modes: normal or accelerated. In the normal mode a car covers 1kilometer in 2 minutes, and consumes 1 liter of fuel. In the accelerated mode a car covers 1 kilometer in 1 minutes, but consumes 2 liters of fuel. The driving mode can be changed at any moment and any number of times.

Your task is to choose a car with minimum price such that Vasya can reach the cinema before the show starts, i.e. not later than in t minutes. Assume that all cars are completely fueled initially.

Input

The first line contains four positive integers n, k, s and t (1 ≤ n ≤ 2·105, 1 ≤ k ≤ 2·105, 2 ≤ s ≤ 109,1 ≤ t ≤ 2·109) — the number of cars at the car rental service, the number of gas stations along the road, the length of the road and the time in which the film starts.

Each of the next n lines contains two positive integers ci and vi (1 ≤ ci, vi ≤ 109) — the price of the i-th car and its fuel tank capacity.

The next line contains k distinct integers g1, g2, ..., gk (1 ≤ gi ≤ s - 1) — the positions of the gas stations on the road in arbitrary order.

Output

Print the minimum rent price of an appropriate car, i.e. such car that Vasya will be able to reach the cinema before the film starts (not later than in t minutes). If there is no appropriate car, print -1.

Examples

input

3 1 8 10
10 8
5 7
11 9
3

output

10

input

2 2 10 18
10 4
20 6
5 3

output

20

Note

In the first sample, Vasya can reach the cinema in time using the first or the third cars, but it would be cheaper to choose the first one. Its price is equal to 10, and the capacity of its fuel tank is 8. Then Vasya can drive to the first gas station in the accelerated mode in 3 minutes, spending 6 liters of fuel. After that he can full the tank and cover 2kilometers in the normal mode in 4 minutes, spending 2 liters of fuel. Finally, he drives in the accelerated mode covering the remaining 3 kilometers in 3 minutes and spending 6 liters of fuel.

题意：有n辆车，路程为s，途中有k个加油站，时间为t；每辆车有价格c，油箱容量v；此时，要求话最少的钱租一辆能够在规定时间里走完全程的车。求最少花多少钱。（加油站免费加满油箱，且不消耗时间）

这道题是自己想出来的，虽然比赛的时候没有思路，第二天还是搞出来了。

思路：二分车辆(0--n-1),对每辆车检验其能否满足条件。

注意：要二分，必须是有序的，所以要先去重，即同价位的车保留容量最大的；第二个，价格高但容量低的车直接抛弃。在这里wa了几次。这两个处理完，二分，然后ac。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;
#define N 200005

struct Car
{
        int p;
        int c;
}car[N];
bool cmp(Car a,Car b)
{
        return a.p<b.p;
}

map<int,int>ma;
int n,k,s,t;
int gas[N];
bool check(int id)
{
        int time=;
        for(int i=;i<=k;i++)
        {
                if(gas[i+]-gas[i]>car[id].c)
                        return ;
                time+=(gas[i+]-gas[i])*;
                time-=min(car[id].c-(gas[i+]-gas[i]),gas[i+]-gas[i]);
                if(time>t)
                        return ;
        }
       // cout<<" time:"<<time<<" ";
        return ;
}
int main()
{
        scanf("%d%d%d%d",&n,&k,&s,&t);
        ma.clear();
        for(int i=;i<n;i++)
        {
                int pp,cc;
                scanf("%d%d",&pp,&cc);
                if(ma.count(pp)==)
                        ma[pp]=cc;
                else if(cc>ma[pp])
                        ma[pp]=cc;
        }
        map<int,int>::iterator it;
        int cnt=;
        for(it=ma.begin();it!=ma.end();it++)
        {
                car[cnt].p=it->first;
                car[cnt++].c=it->second;
        }
        sort(car,car+cnt,cmp);
        int cnta=,maxn=;
        for(int i=;i<cnt;i++)
        {
                if(car[i].c>maxn)
                {
                        maxn=car[i].c;
                        car[cnta++]=car[i];
                }
        }
        for(int i=;i<=k;i++)
                scanf("%d",&gas[i]);
        gas[]=;
        gas[k+]=s;
        sort(gas+,gas+k+);
        int l=,r=cnta-,res=-;
        while(l<=r)
        {
                int mid=(r+l)>>;
                //cout<<mid<<" "<<check(mid)<<endl;
                if(check(mid))
                {
                        res=mid;
                        r=mid-;
                }
                else
                        l=mid+;
        }
        if(res<)
                printf("%d\n",res);
        else
                printf("%d\n",car[res].p);
        return ;
}