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;
int main() {
  long long a;
  cin >> a;
  vector<pair<long long, long long> > vec;
  for (long long i = 0; i < a; i++) {
    long long g, h;
    cin >> g >> h;
    pair<long long, long long> par;
    par.first = g;
    par.second = h;
    vec.push_back(par);
  }
  sort(vec.begin(), vec.end());
  long long tmp = 0;
  vector<long long> estado(a);
  long long compartir = 0;
  for (long long i = 0; i < vec.size(); i++) {
    compartir = max(compartir, estado[i]);
    estado[i] = compartir;
    long long h = vec.size() - 1, d = 0;
    while (h - d > 1) {
      long long m = (h + d) / 2;
      if (vec[m].first > vec[i].second) {
        h = m;
      } else if (vec[m].first <= vec[i].second) {
        d = m;
      }
    }
    tmp = h;
    if (vec[tmp].first > vec[i].second) {
      estado[tmp] = max(estado[tmp], estado[i] + 1);
    }
  }
  cout << compartir + 1 << endl;
}
