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 ri each (1 ≤ li ≤ ri ≤ 109).

Output

Print the maximal number of orders that can be accepted.

Examples

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 3
6 8


Output

2
import java.io.*;
import java.util.*;
public class B597 implements Comparable<B597>
{
    int st,end;
    public B597(int s,int e)
    {
        this.st=s;
        this.end=e;
    }
    public int compareTo(B597 ob)
    {
        return Integer.compare(this.end,ob.end);
    }
    public static void main(String args[])throws IOException
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        B597 arr[]=new B597[n];
        HashMap<Integer,Integer> map=new HashMap<>();
        List<Integer> list=new ArrayList<>();
        for(int i=0;i<n;i++)
        {
            int start=sc.nextInt();
            int end=sc.nextInt();
            arr[i]=new B597(start,end);
            if(!map.containsKey(end))
            {
                map.put(end,0);
                list.add(end);
            }
        }
        Arrays.sort(arr);
        Collections.sort(list);
        int dp[]=new int[n];
        dp[0]=1;
        map.put(arr[0].end,1);
        for(int i=1;i<n;i++)
        {
            int ind=Collections.binarySearch(list,arr[i].st);
            if(ind<0)
            {
                ind=Math.abs(ind)-2;
            }
            else
                ind--;
            if(ind<0)
            {
                int max=(int)Math.max(dp[i-1],Math.max(1,map.get(arr[i].end)));
                dp[i]=max;
                map.put(arr[i].end,max);
            }
            else
            {
                int max=(int)Math.max(dp[i-1],Math.max(map.get(list.get(ind))+1,map.get(arr[i].end)));
                dp[i]=max;
                map.put(arr[i].end,max);
            }
        }
        System.out.println(dp[n-1]);
    }
}
