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;
struct activity {
  int st, ft;
  activity() {}
  activity(int st, int ft) {
    this->st = st;
    this->ft = ft;
  }
};
bool compare(const activity& a1, const activity& a2) {
  if (a1.ft < a2.ft)
    return true;
  else if (a1.ft > a2.ft)
    return false;
  return a1.st < a2.st;
}
int main() {
  int n;
  scanf("%d\n", &n);
  vector<activity> vec(n);
  for (int i = 0; i < n; i++) scanf("%d %d\n", &vec[i].st, &vec[i].ft);
  sort(vec.begin(), vec.end(), compare);
  int prev = 0;
  int res = 1;
  for (int i = 1; i < n; i++) {
    if (vec[i].st > vec[prev].ft) {
      res++;
      prev = i;
    }
  }
  cout << res << endl;
}
