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;
const int N = 1e+6 + 1;
int t[N];
void upd(int i, int x) {
  for (; i < N; i |= i + 1) t[i] = max(t[i], x);
}
int max(int i) {
  int res = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) res = max(t[i], res);
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> l(n), r(n);
  vector<pair<int, int>> b(n);
  vector<int> a(2 * n);
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    a[2 * i] = l[i], a[2 * i + 1] = r[i];
  }
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  for (int i = 0; i < n; i++) {
    b[i].first = lower_bound(a.begin(), a.end(), l[i]) - a.begin();
    b[i].second = lower_bound(a.begin(), a.end(), r[i]) - a.begin();
  }
  sort(b.begin(), b.end(), [](pair<int, int> &x, pair<int, int> &y) {
    return x.second < y.second;
  });
  for (int i = 0; i < n; i++) l[i] = b[i].first, r[i] = b[i].second;
  for (int i = 0; i < n; i++) upd(r[i], max(l[i] - 1) + 1);
  cout << max(N - 1);
}
