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.File;
import java.io.FileNotFoundException;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;


public class CountOrder {
	static class Pair{
		int start;
		int end;
		Pair(int start, int end) {
			this.start = start;
			this.end = end;
		}
	}
	public int countOrder(Pair[] array) {
		Arrays.sort(array, new Comparator<Pair>(){
			@Override
			public int compare(Pair p1, Pair p2) {
				return p1.end - p2.end;
			}
		});
		//use M[i] to store the max orders from index 0 to i
		int[] M= new int[array.length];
		M[0] = 1;
		for(int i = 1; i < array.length; i++) {
			M[i] = M[i - 1];
			int j = binarySearch(array, 0, i - 1, array[i].start);
			if(array[j].end < array[i].start) {
				M[i] = Math.max(M[i], M[j] + 1);
			}
		}
		return M[array.length - 1];
	}
	private int binarySearch(Pair[] array, int i, int j, int target) {
		while(i < j - 1) {
			int mid = i + (j - i) / 2;
			if(array[mid].end >= target) {
				j = mid - 1;
			}
			else {
				i = mid;
			}
		}
		return array[j].end < target? j : i;
	}
	public static void main(String[] args) throws FileNotFoundException  {
		CountOrder solu = new CountOrder();
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		Pair[] array = new Pair[n];
		for(int i = 0; i < n; i++) {
			int start = sc.nextInt();
			int end = sc.nextInt();
			Pair cur = new Pair(start, end);
			array[i] = cur;
		}
		System.out.println(solu.countOrder(array));
	} 
}
