#include <bits/stdc++.h>
using namespace std;
int n, dp[500005];
pair<int, int> p[500005];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d%d", &p[i].first, &p[i].second);
  sort(p, p + n);
  for (int i = 0; i < n; ++i) {
    if (i) dp[i] = max(dp[i], dp[i - 1]);
    int ub = upper_bound(p + i + 1, p + n, make_pair(p[i].second + 1, 0)) - p;
    dp[ub] = max(dp[ub], dp[i] + 1);
  }
  printf("%d\n", max(dp[n], dp[n - 1]));
}

### Prompt

Generate 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;
int n, dp[500005];
pair<int, int> p[500005];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d%d", &p[i].first, &p[i].second);
  sort(p, p + n);
  for (int i = 0; i < n; ++i) {
    if (i) dp[i] = max(dp[i], dp[i - 1]);
    int ub = upper_bound(p + i + 1, p + n, make_pair(p[i].second + 1, 0)) - p;
    dp[ub] = max(dp[ub], dp[i] + 1);
  }
  printf("%d\n", max(dp[n], dp[n - 1]));
}

```