Restaurant

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 li and the finish time ri (li ≤ ri).

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 li and rieach (1 ≤ li ≤ ri ≤ 109).

Output

Print the maximal number of orders that can be accepted.

Sample Input

Input

2
7 11
4 7

Output

1

Input

5
1 2
2 3
3 4
4 5
5 6

Output

3

Input

6
4 8
1 5
4 7
2 5
1 36 8

Output

2

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;
}