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;
bool IsPrime(long long int x) {
  bool isPrime = true;
  for (long long int a = 2; a * a <= x; a++) {
    if (x % a == 0) return false;
  }
  return true;
}
long long int floorSqrt(long long int x) {
  if (x == 0 || x == 1) return x;
  long long int start = 1, end = x, ans;
  while (start <= end) {
    long long int mid = (start + end) / 2;
    if (mid * mid == x) return mid;
    if (mid * mid < x) {
      start = mid + 1;
      ans = mid;
    } else
      end = mid - 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i, n, x = 0, k = 0;
  cin >> n;
  set<pair<long long int, long long int> > se1;
  for (i = 0; i < n; i++) {
    pair<long long int, long long int> p1;
    cin >> p1.second >> p1.first;
    se1.insert(p1);
  }
  for (auto j = se1.begin(); j != se1.end(); j++) {
    pair<long long int, long long int> p = *j;
    if (p.second > k) x++, k = p.first;
  }
  cout << x;
  return 0;
}
