#include <bits/stdc++.h>
using namespace std;
struct point {
  int x;
  int y;
};
point dates[500005];
bool cmp(point a, point b) {
  if (a.y < b.y) {
    return true;
  }
  if (a.y == b.y) {
    return a.x < b.x;
  }
  return false;
}
int main() {
  cin.sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> dates[i].x >> dates[i].y;
  }
  sort(dates, dates + n, cmp);
  int var = dates[0].y;
  int ans = 1;
  for (int i = 1; i < n; i++) {
    if (dates[i].x > var) {
      ans++;
      var = dates[i].y;
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
struct point {
  int x;
  int y;
};
point dates[500005];
bool cmp(point a, point b) {
  if (a.y < b.y) {
    return true;
  }
  if (a.y == b.y) {
    return a.x < b.x;
  }
  return false;
}
int main() {
  cin.sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> dates[i].x >> dates[i].y;
  }
  sort(dates, dates + n, cmp);
  int var = dates[0].y;
  int ans = 1;
  for (int i = 1; i < n; i++) {
    if (dates[i].x > var) {
      ans++;
      var = dates[i].y;
    }
  }
  cout << ans << '\n';
  return 0;
}

```