Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.

The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.

The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.

The next n lines each contain two space-separated integers f_i and t_i (1 ≤ f_i ≤ s, 1 ≤ t_i ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.

Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.

In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:

1. Move to floor 5: takes 2 seconds.

2. Pick up passenger 3.

3. Move to floor 3: takes 2 seconds.

4. Wait for passenger 2 to arrive: takes 4 seconds.

5. Pick up passenger 2.

6. Go to floor 2: takes 1 second.

7. Pick up passenger 1.

8. Go to floor 0: takes 2 seconds.

This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.

题意就是电梯接乘客，这个电梯只能从上往下走，给的数据是在第几层有乘客，乘客到达该层的时间，乘客上电梯的时间忽略不计，要求输出花费的最短时间是多少。

我的理解就是单独把某层的时间拿出来算一下，从该层到最底层花的时间+等待乘客的时间和总的层数花的时间比较一下，找出max，然后其他层也是这样，和上一组楼层的max比较一下。其实总的想一下，就是把楼层数+等待乘客的时间找出max，然后和总的楼层数比较一下就可以了。(最近好多题都用到这种思想了，不知道这种思想叫啥，但是感觉有时候挺好用的，就是有时用这种思路写题会T，超时了耶。。。)

代码:

#include<bits/stdc++.h>

using namespace std;

int main(){

    int n,m,f,t;

    int maxx=-;

    while(~scanf("%d%d",&n,&m)){

        for(int i=;i<n;i++){

            scanf("%d%d",&f,&t);

            if(f+t>maxx)

            maxx=max(f+t,maxx);

        }

        if(maxx<m)

            printf("%d\n",m);

        else

            printf("%d\n",maxx);

    }

    return ;

}