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 gcd(long long a, long long b) { return (b ? gcd(b, a % b) : a); }
bool isPrime(long long n) {
  if (n == 1) return false;
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long power(long long x, unsigned long long y, long long p = 1000000007) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long digit(long long n) {
  long long ans = 0;
  while (n) {
    ans++;
    n /= 10;
  }
  return ans;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
vector<long long> sieve(long long n) {
  long long *arr = new long long[n + 1]();
  vector<long long> vect;
  for (long long i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (long long j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].second;
    cin >> v[i].first;
  }
  sort(v.begin(), v.end());
  long long end = 0;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    if (v[i].second > end) {
      ans++;
      end = v[i].first;
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base ::sync_with_stdio(false), cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
