#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 1e5 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, l, r;
  cin >> n;
  vector<pair<long long, long long> > v;
  for (int i = 1; i <= n; i++) {
    cin >> l >> r;
    v.push_back(make_pair(l, -i));
    v.push_back(make_pair(r, i));
  }
  int c = 0;
  sort(v.begin(), v.end());
  set<long long> op;
  for (int i = 0; i < 2 * n; i++) {
    if (v[i].second < 0) {
      op.insert(-v[i].second);
    } else if (op.count(v[i].second)) {
      op.clear();
      c++;
    }
  }
  cout << c << endl;
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 1e5 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, l, r;
  cin >> n;
  vector<pair<long long, long long> > v;
  for (int i = 1; i <= n; i++) {
    cin >> l >> r;
    v.push_back(make_pair(l, -i));
    v.push_back(make_pair(r, i));
  }
  int c = 0;
  sort(v.begin(), v.end());
  set<long long> op;
  for (int i = 0; i < 2 * n; i++) {
    if (v[i].second < 0) {
      op.insert(-v[i].second);
    } else if (op.count(v[i].second)) {
      op.clear();
      c++;
    }
  }
  cout << c << endl;
  return 0;
}

```