CodeForces - 608A-Saitama Destroys Hotel（模拟）

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 sat 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.

Input

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 fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.

Output

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

Examples

Input

3 7
2 1
3 8
5 2

Output

11

Input

5 10
2 77
3 33
8 21
9 12
10 64

Output

79

题解：模拟就好了，每次比较他的到达时间和电梯到达时间谁大，谁大就更新谁，最后加上最后一项的楼层数

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

struct node {
	int a,b;

} p[1005];

bool cmp(node x,node y) {
	return x.a>y.a;
}
int main() {
	int n,s;
	cin>>n>>s;
	for(int t=0; t<n; t++) {
		scanf("%d%d",&p[t].a,&p[t].b);
	}
	sort(p,p+n,cmp);
	int sum=max(s-p[0].a,p[0].b);
	for(int t=1; t<n; t++) {
		sum=max(sum+p[t-1].a-p[t].a,p[t].b);
	}
	cout<<sum+p[n-1].a<<endl;
	return 0;
}