#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
  int n, l, r, ri = 0;
  cin >> n;
  vector<pair<int, int> > tm;
  for (long long i = 0; i < n; i++) {
    cin >> l >> r;
    tm.push_back(make_pair(l, r));
  }
  sort(tm.begin(), tm.end(), cmp);
  int cont = 1;
  ri = tm[0].second;
  cout << endl;
  for (long long i = 1; i < tm.size(); i++) {
    if (tm[i].first <= ri)
      continue;
    else {
      ri = tm[i].second;
      cont++;
    }
  }
  cout << cont << endl;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 = 1000000007;
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
  int n, l, r, ri = 0;
  cin >> n;
  vector<pair<int, int> > tm;
  for (long long i = 0; i < n; i++) {
    cin >> l >> r;
    tm.push_back(make_pair(l, r));
  }
  sort(tm.begin(), tm.end(), cmp);
  int cont = 1;
  ri = tm[0].second;
  cout << endl;
  for (long long i = 1; i < tm.size(); i++) {
    if (tm[i].first <= ri)
      continue;
    else {
      ri = tm[i].second;
      cont++;
    }
  }
  cout << cont << endl;
}

```