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
#include <bits/stdc++.h>
using namespace std;
typedef struct {
  int number;
  bool is_begin;
  int time;
} times;
bool myfunction(times i, times j) { return (i.time < j.time); }
int main() {
  int n;
  cin >> n;
  vector<times> t;
  for (int i = 0; i < n; i++) {
    int begin, end;
    cin >> begin;
    cin >> end;
    times a, b;
    a.is_begin = true;
    b.is_begin = false;
    a.number = i;
    b.number = i;
    a.time = begin;
    b.time = end;
    t.push_back(a);
    t.push_back(b);
  }
  stable_sort(t.begin(), t.end(), myfunction);
  set<int> beginned;
  int result = 0;
  int last = -1;
  for (int i = 0; i < 2 * n; i++) {
    if (t[i].is_begin && t[i].time > last) {
      beginned.insert(t[i].number);
    } else if (!t[i].is_begin && beginned.find(t[i].number) != beginned.end()) {
      result++;
      last = t[i].time;
      beginned.clear();
    }
  }
  cout << result << "\n";
  return 0;
}
