codeforces845C（stl）-白红宇

codeforces845C（stl）

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发布时间：2019-04-26

本文共 1372 字，大约阅读时间需要 4 分钟。

Polycarp is a great fan of television.

He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment li and ends at moment ri.

Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.

Polycarp wants to check out all n shows. Are two TVs enough to do so?

Input

The first line contains one integer n (1 ≤ n ≤ 2·105) — the number of shows.

Each of the next n lines contains two integers li and ri (0 ≤ li < ri ≤ 109) — starting and ending time of i-th show.

Output

If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).

Examples

input

3 1 2 2 3 4 5

output

YES

input

4 1 2 2 3 2 3 1 2

output

NO

题意：给你n个区间，问你将这些区间分成不相交的个数是否小于等于2（注意：不能有重点）。

思路：先对l排序，用multiset维护不相交区间个数。

代码：

#include
      
       using namespace std;const int maxn=2e5+5;struct node{
           int l;    int r;}a[maxn];bool cmp(const node &x,const node &y){
           return x.l

s; for(int i=0;i

2) puts("NO"); else puts("YES"); } return 0;}

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