CF2.C（二分贪心）

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 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 c_i and v_i — 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 v_i. 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 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 c_i and v_i (1 ≤ c_i, v_i ≤ 10⁹) — the price of the i-th car and its fuel tank capacity.

The next line contains k distinct integers g₁, g₂, ..., g_k (1 ≤ g_i ≤ 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 2 kilometers 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，每辆车有租金和邮箱的容量，路上有k个加油站，道路总长s，要求在t时间内到达终点，问选哪辆车可以以最少的租金在规定时间内到达，加油站加油不要钱，最后一行给出加油站的位置。

代码：

 //直接找会在第31组样例TE,卡了好久。二分油箱，找到一个最小的油箱容量能够走完全程，汽车排序后找到第一个大于等于该容量的汽车即可
 //二分啊，每次都想不清楚，写不对！
 #include<bits\stdc++.h>
 using namespace std;
 struct Lu
 {
     int r,v;
 }L[];
 int com[];
 int n,k,s,t;
 bool cmp1(Lu x,Lu y)
 {
     return x.r<y.r;
 }
 bool cmp2(int x,int y)
 {
     return x<y;
 }
 int check(long long mid)
 {
     int m=,now=,st=t;
     while(m<k)
     {
         if(com[m]-now>mid||com[m]-now>st)
             return ;
         long long tem=mid-(com[m]-now);//剩余多少油就说明可以有多少路高速行驶
         if(tem>com[m]-now)
             tem=com[m]-now;
         st-=(tem+*(com[m]-now-tem));
         now=com[m];
         m++;
     }
     if(s-now>mid||s-now>st)
         return ;
     long long tem=mid-(s-now);
     if(tem>s-now)
         tem=s-now;
     st-=(tem+*(s-now-tem));
     if(st<) return ;
     return ;
 }
 int main()
 {
     scanf("%d%d%d%d",&n,&k,&s,&t);
     for(int i=;i<n;i++)
         scanf("%d%d",&L[i].r,&L[i].v);
     for(int i=;i<k;i++)
         scanf("%d",&com[i]);
     sort(L,L+n,cmp1);
     sort(com,com+k,cmp2);
     long long lef=,rig=,mid;
     while(lef<=rig)
     {
         mid=(lef+rig)/;
         if(check(mid))
             rig=mid-;
         else lef=mid+;
     }
    // cout<<rig+1<<endl;
     int ans=-;
     for(int i=;i<n;i++)
     if(L[i].v>=rig+)
     {
         ans=L[i].r;
         break;
     }
     printf("%d\n",ans);
     return ;
 }