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 INF = 2000000000;
int t[4000400];
map<int, int> m;
pair<int, int> l[500500];
int fsum(int l, int r, int lv, int rv, int v) {
  if (l > rv || r < lv) return 0;
  if (lv >= l && rv <= r) return t[v];
  return max(fsum(l, r, lv, (lv + rv) / 2, v * 2),
             fsum(l, r, (lv + rv) / 2 + 1, rv, v * 2 + 1));
}
void modify(int v, int d) {
  t[v] = max(t[v], d);
  v /= 2;
  for (; v > 0; v /= 2) t[v] = max(t[v * 2], t[v * 2 + 1]);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> l[i].first >> l[i].second;
    m[l[i].first] = 0;
    m[l[i].second] = 0;
  }
  int k = 0;
  for (map<int, int>::iterator it = m.begin(); it != m.end(); it++)
    it->second = ++k;
  for (int i = 0; i < n; i++) {
    l[i].first = m[l[i].first];
    l[i].second = m[l[i].second];
  }
  sort(l, l + n);
  int z = 1, ans = 0;
  while (z < k) z *= 2;
  for (int i = 0; i < n; i++) {
    int s = fsum(1, l[i].first - 1, 1, z, 1);
    modify(z + l[i].second - 1, s + 1);
    ans = max(ans, s + 1);
  }
  cout << ans << endl;
  return 0;
}
