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;
const int N = 5e5 + 55;
vector<int> vec[N * 2];
map<int, int> mp;
int dp[N * 2];
int s[N];
int e[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> temp;
  for (int i = 0; i < n; i++) {
    cin >> s[i] >> e[i];
    temp.push_back(s[i]);
    temp.push_back(e[i]);
  }
  sort(temp.begin(), temp.end());
  int cnt = 1;
  for (auto v : temp) {
    if (mp[v] == 0) mp[v] = cnt++;
  }
  for (int i = 0; i < n; i++) {
    vec[mp[e[i]]].push_back(mp[s[i]]);
  }
  int ans = 0;
  for (int i = 1; i < cnt; i++) {
    dp[i] = dp[i - 1];
    for (auto v : vec[i]) {
      dp[i] = max(dp[i], dp[v - 1] + 1);
    }
    ans = max(ans, dp[i]);
  }
  cout << ans << endl;
  return 0;
}
