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 fast_exp(long long int base, long long int exp1) {
  long long int res = 1;
  while (exp1 > 0) {
    if (exp1 & 1) res = (res * base) % 1000000007;
    base = (base * base) % 1000000007;
    exp1 /= 2;
  }
  return res % 1000000007;
}
long long int pr[1000001] = {0};
void isprime() {
  pr[0] = 1;
  pr[1] = 1;
  for (long long int a = 2; a < 1000001; a++) {
    if (!pr[a]) {
      for (long long int b = a * 2; b < 1000001; b += a) {
        pr[b] = 1;
      }
    }
  }
}
long long int comp = 1, tim = 0;
struct vertex {
  vector<long long int> adj;
  long long int vis = 0;
  long long int parent = -1;
  long long int dist = 0;
  long long int component = 0;
  long long int in = 0;
  long long int out = 0;
  long long int low = 0;
  long long int child = 0;
  long long int indeg = 0;
  long long int outdeg = 0;
  bool AP = false;
};
long long int n;
vector<pair<long long int, long long int> > v;
bool sortsec(pair<long long int, long long int> a,
             pair<long long int, long long int> b) {
  return a.second < b.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long int a = 0; a < n; a++) {
    long long int b, c;
    cin >> b >> c;
    v.push_back(make_pair(b, c));
  }
  sort(v.begin(), v.end(), sortsec);
  long long int last = 0, ans = 0;
  for (long long int a = 0; a < n; a++) {
    if (v[a].first > last) {
      last = v[a].second;
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
