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(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> t(n + 1);
  for (long long i = 1; i <= n; i++) cin >> t[i].first >> t[i].second;
  sort(t.begin() + 1, t.end());
  long long ans = 1, prev_l = t[1].first, prev_r = t[1].second;
  for (long long i = 2; i <= n;) {
    if (t[i].first == prev_l) {
      long long j = i;
      while (t[j].first == prev_l) j++;
      i = j;
    } else if (t[i].first > prev_l && t[i].second < prev_r)
      prev_l = t[i].first, prev_r = t[i].second, i++;
    else if (t[i].second >= prev_r && t[i].first <= prev_r)
      i++;
    else if (t[i].first > prev_r) {
      ans++;
      prev_l = t[i].first;
      prev_r = t[i].second;
      i++;
    }
  }
  cout << ans << "\n";
  return 0;
}
