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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Random;
import java.util.Set;
import java.util.Stack;

public class test {
	static class Order {
		int start, finish;
		public Order(int s, int f) {
			start  = s;
			finish = f;
		}
	}
	
	public static void main(String[] args) throws InterruptedException {
		//new careercup().run();
		//new CC().run();
		try{
			BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			int n = Integer.parseInt(br.readLine());
			PriorityQueue<Order> pq = new PriorityQueue<>((o1, o2) -> (Integer.compare(o1.finish, o2.finish)));
			for(int i=0; i<n; i++) {
				String[] s = br.readLine().split(" ");
				pq.offer(new Order(Integer.parseInt(s[0]), Integer.parseInt(s[1])));
			}	
			int lastFinish = 0, count = 0;
			while(!pq.isEmpty()) {
				if(pq.peek().start > lastFinish) {
					lastFinish = pq.peek().finish;
					count++;
				}
				pq.poll();
			}
			System.out.println(count);
		}catch(IOException io){
			io.printStackTrace();
		}
	}
}
