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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
const int mxN = 5e5;
vector<pair<int, int> > a;
int store[mxN], n;
int dp(int i) {
  if (i >= n) return 0;
  if (store[i] != -1) return store[i];
  int lb = i, rb = n, val = a[i].second;
  while (lb < rb) {
    int mb = (lb + rb) / 2;
    if (a[mb].first <= val) {
      lb = mb + 1;
    } else {
      rb = mb;
    }
  }
  return store[i] = max(dp(i + 1), 1 + dp(lb));
}
int main() {
  cin >> n;
  a = vector<pair<int, int> >(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first >> a[i].second;
  }
  sort(a.begin(), a.end());
  a.push_back(make_pair(2e9, 2e9));
  memset(store, -1, sizeof(store));
  int best = 0;
  for (int i = 0; i < n; ++i) {
    best = max(best, dp(i));
  }
  cout << best << "\n";
}
