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;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int bin(long int a[], long int low, long int high, long d, long day[]) {
  if (low > high) return -1;
  long mid = (low + high) / 2;
  if (a[mid] > (a[low] + d) && (a[mid - 1] <= (a[low] + d) || day[mid - 1]) &&
      day[mid] == 0)
    return mid;
  else if (a[mid] > (a[low] + d) && day[mid] == 0)
    return bin(a, low, mid - 1, d, day);
  else
    return bin(a, mid + 1, high, d - a[mid + 1] + a[low], day);
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) return (power((a * a) % 1000000007, b / 2)) % 1000000007;
  return ((a % 1000000007) * (power((a * a) % 1000000007, b / 2)) %
          1000000007) %
         1000000007;
}
int main() {
  int t = 1;
  while (t--) {
    long long int n, i, cs = 0, count = 0, j, a, b;
    cin >> n;
    vector<pair<long long int, long long int> > v;
    for (long long int i = 0; i < (n); i++) {
      cin >> a >> b;
      v.push_back(make_pair(b, a));
    }
    sort((v).begin(), (v).end());
    i = 0;
    while (i < n) {
      j = i + 1;
      while (j < n && v[i].first == v[j].first) j++;
      i = j - 1;
      if (v[i].second > cs) {
        cs = v[i].first;
        count++;
      }
      i++;
    }
    cout << count << "\n";
  }
  return 0;
}
