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;
void solve() {
  int n;
  cin >> n;
  vector<pair<long long, long long> > a;
  for (int i = 0; i < n; i++) {
    long long s, e;
    cin >> s >> e;
    a.push_back(make_pair(s, e));
  }
  sort(a.begin(), a.end());
  int res = 0;
  map<long long, int> best;
  best[0] = 0;
  for (int i = 0; i < n; i++) {
    long long st = a[i].first;
    long long en = a[i].second;
    auto lower = best.lower_bound(st);
    lower--;
    int cb = lower->second + 1;
    best[en] = cb;
    auto it = best.lower_bound(en);
    ++it;
    while (it != best.end()) {
      if (it->second < cb) {
        it = best.erase(it);
      } else {
        break;
      }
    }
    res = max(res, cb);
  }
  cout << res << endl;
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
