#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > seg;
int cur_end = -1, r = 0;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0, a, b; i < n; ++i) {
    scanf("%d%d", &a, &b);
    seg.push_back({b, a});
  }
  sort(seg.begin(), seg.end());
  for (int i = 0; i < n; ++i) {
    if (seg[i].second > cur_end) r++, cur_end = seg[i].first;
  }
  cout << r;
}

### Prompt

In Cpp, your task is to solve 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;
vector<pair<int, int> > seg;
int cur_end = -1, r = 0;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0, a, b; i < n; ++i) {
    scanf("%d%d", &a, &b);
    seg.push_back({b, a});
  }
  sort(seg.begin(), seg.end());
  for (int i = 0; i < n; ++i) {
    if (seg[i].second > cur_end) r++, cur_end = seg[i].first;
  }
  cout << r;
}

```