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 = 500010;
int ans[N], arr[N], MAX[N];
long long power_mod(long long x, long long y) {
  x %= 1000000007;
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = (res * x) % 1000000007;
    }
    y /= 2;
    x = (x * x) % 1000000007;
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> v(N);
  for (int i = 1; i <= n; i++) {
    cin >> v[i].first >> v[i].second;
    ans[i] = MAX[i] = 1;
  }
  sort(v.begin() + 1, v.begin() + n + 1);
  arr[n + 1] = v[n].second + 1;
  for (int i = 1; i <= n; i++) {
    arr[i] = v[i].first;
  }
  int ma = 0;
  for (int i = n; i >= 1; i--) {
    int up = upper_bound(arr + 1, arr + 2 + n, v[i].second) - arr;
    int p = ans[i];
    ans[i] = max(ans[i], ans[i + 1]);
    ans[i] = max(ans[i], p + ans[up]);
  }
  int res = *max_element(ans, ans + n + 1);
  cout << res << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}
