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
#include <bits/stdc++.h>
using namespace std;
const int N = 500010;
int n, p, q, b[N + N + 10], h[N + N + 10], dp[N + N + 10];
pair<int, int> a[N];
vector<int> V[N + N + 10];
void spusti() {
  sort(b + 1, b + p + 1);
  h[1] = b[1];
  for (int i = 2; i <= p; i++)
    if (b[i] != b[i - 1]) h[++q] = b[i];
  for (int i = 1; i <= n; i++) {
    a[i].first = lower_bound(h + 1, h + q + 1, a[i].first) - h;
    a[i].second = lower_bound(h + 1, h + q + 1, a[i].second) - h;
  }
}
int main() {
  scanf("%d", &n);
  p = 0;
  q = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].first, &a[i].second);
    b[++p] = a[i].first;
    b[++p] = a[i].second;
  }
  spusti();
  for (int i = 1; i <= n; i++) {
    V[a[i].second].push_back(i);
  }
  dp[0] = 0;
  for (int i = 1; i <= 2 * N; i++) dp[i] = 0;
  for (int i = 1; i <= 2 * N; i++) {
    if (!V[i].empty())
      for (int j = 0; j < V[i].size(); j++)
        dp[i] = max(dp[i], dp[a[V[i][j]].first - 1] + 1);
    dp[i] = max(dp[i], dp[i - 1]);
  }
  printf("%d", dp[2 * N]);
  return 0;
}
