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 compare(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second < b.second;
}
int main() {
  int num;
  cin >> num;
  vector<pair<int, int>> orders(num);
  map<int, int> compress;
  for (int i = 0; i < num; i++) {
    int l, r;
    cin >> l >> r;
    compress[l] = 0;
    compress[r] = 0;
    orders[i] = {l, r};
  }
  int n = 0;
  for (auto& k : compress) k.second = ++n;
  vector<vector<int>> endings(n + 5);
  vector<int> dp(n + 5);
  for (auto& k : orders) {
    endings[compress[k.second]].push_back(compress[k.first]);
  }
  for (int i = 1; i <= n; i++) {
    int mo = dp[i - 1];
    for (int k : endings[i]) mo = max(mo, dp[k - 1] + 1);
    dp[i] = mo;
  }
  cout << dp[n] << endl;
}
