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;
map<long long, int> mp;
pair<int, int> P[600020];
vector<int> vec;
int res[12000020];
int now = 0, ans = 0;
int n, i;
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &P[i].second, &P[i].first);
    vec.push_back(P[i].first);
    vec.push_back(P[i].second);
  }
  sort(vec.begin(), vec.end());
  int st = -1;
  for (int i = 0; i < vec.size(); i++) {
    if (vec[i] > st) {
      st = vec[i];
      now++;
    }
    mp[vec[i]] = now;
  }
  sort(P + 1, P + n + 1);
  int r = 1;
  for (int i = 1; i <= n; i++) swap(P[i].first, P[i].second);
  for (int i = 1; i <= n; i++) {
    P[i].first = mp[P[i].first];
    P[i].second = mp[P[i].second];
    while (r < P[i].second) {
      res[r] = max(res[r - 1], res[r]);
      r++;
    }
    res[P[i].second] = max(res[P[i].second], res[P[i].first - 1] + 1);
    ans = max(ans, res[P[i].second]);
  }
  cout << ans << endl;
}
