Restaurant

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本文介绍了一个算法案例，解决餐厅在面对多个连续时间段预订请求时如何最大化接受预订数量的问题。通过对比不同预订的时间段，该算法确保了任何两个接受的预订时段不发生重叠。

I - Restaurant

Time Limit:4000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

A restaurant received n orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the i-th order is characterized by two time values — the start time l_i and the finish time r_i (l_i ≤ r_i).

Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept?

No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both.

Input

The first line contains integer number n (1 ≤ n ≤ 5·10⁵) — number of orders. The following n lines contain integer values l_i and r_ieach (1 ≤ l_i ≤ r_i ≤ 10⁹).

Output

Print the maximal number of orders that can be accepted.

Sample Input

Input

2
7 11
4 7

Output

Input

Output

Input

Output

AC代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
	__int64 u;
	__int64 v;
} a[1100002];
bool cmp(node A,node B)
{
	return A.v <B.v ;
}
int main()
{
	__int64 n,i;
	while(scanf("%I64d",&n)!=EOF)
	{
		for(i=0;i<n;i++)
	{
		scanf("%I64d%I64d",&a[i].u ,&a[i].v );
	}
	sort(a,a+n,cmp);
	__int64 sum=1;
	__int64 temp=a[0].v;
	for(i=1;i<n;i++)
	{
		if(a[i].u>temp)
		{
			temp=a[i].v;
		sum++;
		}
		
	}
    printf("%I64d\n",sum);
	}
	
	return 0; 
}