Codeforces Round #380 (Div. 2,Based on Technocup 2017 - Elimination Round 2)C - Road to Cine//

本文探讨了一个有趣的问题：如何选择最优的租车方案以便及时到达目的地。在考虑了不同车辆的价格、油箱容量、行驶速度及沿途加油站位置等因素后，采用二分查找算法确定了最低成本的可行租车选项。

Vasya is currently at a car rental service, and he wants to reach cinema. The film he has bought a ticket for starts in t minutes. 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 1 kilometer in 2minutes, 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 tminutes. 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 3minutes, spending 6 liters of fuel. After that he can full the tank and cover 2 kilometers in the normal mode in 4 minutes, spending 2liters of fuel. Finally, he drives in the accelerated mode covering the remaining 3 kilometers in 3 minutes and spending 6 liters of fuel.

二分1到10^9，先找到符合时间的油箱的容量的最小值，然后通过最小值来找输入的数据中，油箱符合条件（大于等于最小值）的车里，最便宜的那辆车，为什么要找最便宜的车呢，因为便宜的车稳啊（手动滑稽）。

#include<stdio.h>
#include<algorithm>
#define inf 0x3f3f3f3f3f3f3f3f
using namespace std;
long long int c[200005],v[200005],g[200005],lenn[200005];
long long int n,k,s,t;
bool judge (long long int x); 
int cmp(long long int a,long long int b)
{
	return a<b;
}
int main()
{
	long long int le,ri,mid,i;
	scanf("%lld %lld %lld %lld",&n,&k,&s,&t);
	for(int i=0;i<n;i++)
	scanf("%lld %lld",&c[i],&v[i]);
	g[0]=0;
	for(i=1;i<=k;i++)
	{
		scanf("%lld",&g[i]);
	}
	g[++k]=s;
	sort(g,g+k+1,cmp);
	for(i=1;i<=k;i++)
	{
		lenn[i-1]=g[i]-g[i-1];
	}
	//for(i=0;i<n;i++)
//	printf("%d\n",judge(v[i]));
	le=1,ri=1000000005;
	while(le<ri)
	{
		mid=(le+ri)/2;
		if(judge(mid))
		{
			ri=mid;
		}
		else
		{
			le=mid+1;
		}
	}
//	printf("%d %d %d\n",le,mid,ri);
	long long int min=inf;
	for(i=0;i<n;i++)
	{
		if(v[i]>=ri)
		{
			if(min>c[i])
			min=c[i];
		}
	}
	if(min==inf)
	printf("-1\n");
	else
	printf("%lld\n",min);
	return 0;
	
}
bool judge(long long int x)
{
	long long int cost=0;
	for(int i=0;i<k;i++)
	{
		if(lenn[i]>x)
		return 0;
		else if(lenn[i]*2<=x)
		{
			cost+=lenn[i];
		}
		else
		{
			cost+=2*lenn[i];
			cost-=(x-lenn[i]);
		}
	}
	if(cost<=t)
	return 1;
	return 0;
}