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.util.*;
public class Restaurant {
    static class Order {
        int start;
        int end;
        Order(int start, int end) {
            this.start = start;
            this.end = end;
        }
        @Override 
        public boolean equals(Object obj) {
            if (obj == this) {
                return true;
            }
            if (!(this instanceof Order)) {
                return false;
            }
            Order another = (Order)obj;
            return this.start == another.start && this.end == another.end;
        }
    }
    private static Scanner sc;
    public static int maxOrder(Order[] array) {
        assert array != null;
        if (array.length <= 1) {
            return array.length;
        }
        Arrays.sort(array, new Comparator<Order>(){
            @Override
            public int compare(Order one, Order two) {
                if (one.end == two.end) {
                    return 0;
                }
                return one.end < two.end ? -1 : 1;
            }
        });
        // dp[i] represents the maximum order by i-th order end time (not necessary including i)
        int[] dp = new int[array.length];
        //initialization;
        dp[0] = 1;
        for (int i = 1; i < array.length; i++) {
            int index = binarySearch(0, i - 1, array, array[i].start);
            if (index != -1) {
                dp[i] = dp[index];
            }
            dp[i] += 1;
            dp[i] = Math.max(dp[i], dp[i - 1]);
        }
        return dp[array.length - 1];
    }
    public static int binarySearch(int left, int right, Order[] array, int target) {
        while (left < right - 1) {
            int mid = left + (right - left) / 2;
            if (array[mid].end >= target) {
                right = mid - 1;
            } else {
                left = mid;
            }
        }
        if (array[right].end < target) {
            return right;
        }else if (array[left].end < target) {
            return left;
        } else {
            return -1;
        }
    }
    public static void main(String[] args) {
        sc = new Scanner(System.in);
        Order[] array = new Order[sc.nextInt()];
        for (int i = 0; i < array.length; i++) {
            array[i] = new Order(sc.nextInt(), sc.nextInt());
        }
        System.out.print(maxOrder(array));
    }
}
