Codeforces Round #398(Div. 2)B. The Queue【贪心+谨慎】这题好尼玛强劲

B. The Queue

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Finally! Vasya have come of age and that means he can finally get a passport! To do it, he needs to visit the passport office, but it's not that simple. There's only one receptionist at the passport office and people can queue up long before it actually opens. Vasya wants to visit the passport office tomorrow.

He knows that the receptionist starts working after t_s minutes have passed after midnight and closes aftert_f minutes have passed after midnight (so that(t_f - 1) is the last minute when the receptionist is still working). The receptionist spends exactlyt minutes on each person in the queue. If the receptionist would stop working withint minutes, he stops serving visitors (other than the one he already serves).

Vasya also knows that exactly n visitors would come tomorrow. For each visitor Vasya knows the point of time when he would come to the passport office. Each visitor queues up and doesn't leave until he was served. If the receptionist is free when a visitor comes (in particular, if the previous visitor was just served and the queue is empty), the receptionist begins to serve the newcomer immediately.

"Reception 1"

For each visitor, the point of time when he would come to the passport office is positive. Vasya can come to the office at the time zero (that is, at midnight) if he needs so, but he can come to the office only at integer points of time. If Vasya arrives at the passport office at the same time with several other visitors, he yields to them and stand in the queue after the last of them.

Vasya wants to come at such point of time that he will be served by the receptionist, and he would spend the minimum possible time in the queue. Help him!

Input

The first line contains three integers: the point of time when the receptionist begins to workt_s, the point of time when the receptionist stops workingt_f and the time the receptionist spends on each visitort. The second line contains one integer n — the amount of visitors (0 ≤ n ≤ 100 000). The third line contains positive integers in non-decreasing order — the points of time when the visitors arrive to the passport office.

All times are set in minutes and do not exceed 10¹²; it is guaranteed thatt_s < t_f. It is also guaranteed that Vasya can arrive at the passport office at such a point of time that he would be served by the receptionist.

Output

Print single non-negative integer — the point of time when Vasya should arrive at the passport office. If Vasya arrives at the passport office at the same time with several other visitors, he yields to them and queues up the last. If there are many answers, you can print any of them.

Examples

Input

10 15 2210 13

Output

12

Input

8 17 343 4 5 8

Output

2

Note

In the first example the first visitor comes exactly at the point of time when the receptionist begins to work, and he is served for two minutes. At 12 minutes after the midnight the receptionist stops serving the first visitor, and if Vasya arrives at this moment, he will be served immediately, because the next visitor would only come at 13 minutes after midnight.

In the second example, Vasya has to come before anyone else to be served. 

题目大意：

一个工作窗口，从ts开放到tf.工作人员很负责，只要到时间一定上岗，到时间一定离开岗位。

处理每一个人的需求需要用k时间。 

现在已知这一天有N个人将要来办理业务，来的时间已经给出。

现在主人公想要等待的时间最短，问什么时刻去最好。

思路：

1、问题的大方向很好考虑：

①要么主人公在所有人之前来，要么主人公在所有人之后来。

②再要么就是在每一个人之前一个单位时间来。

以上是所有可能的最优时间。

那么我们需要对其进行逐一判断。

2、需要注意几个点：

①这N个人来的时间都是此起彼伏的，有可能早来，当然，也有可能迟到。

②有可能一个人都没有，就是N==0的情况。

③我萌一定要在结束工作之前，办理业务。

④在最开始来的时候，时间不能为负。

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细节很多的一个题，坑点很多...................

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<map>using namespace std;#define ll __int64ll a[100060];int main(){    ll l,r,t;    while(~scanf("%I64d%I64d%I64d",&l,&r,&t))    {        map<ll,int >s;        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%I64d",&a[i]);            s[a[i]]=1;        }        sort(a+1,a+1+n);        ll wait=0x3f3f3f3f;        ll output=0x3f3f3f3f;        ll now;        ll preend;        if(a[1]-1>=0)output=a[1]-1,wait=abs(output-l);//所有人之前来        if(l<a[1])output=l,wait=0;        for(int i=1;i<=n;i++)        {            ll tmpoutput=output;            if(i==1)            {                if(a[i]<l)now=l;                else now=a[i];                preend=now+t;            }            else            {                if(preend<a[i])                {                    output=preend;                    wait=0;                }                else                {                    if(preend-(a[i]-1)<wait)                    {                        wait=preend-(a[i]-1);                        output=a[i]-1;                    }                    now=preend;                    preend=now+t;                }            }            if(output+t>=r)output=tmpoutput;        }        if(preend+t<=r)output=preend;//所有人之后来        if(n==0)output=l;        printf("%I64d\n",output);    }}