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 SIZE = 400 + 1;
const long long int mod = 1e9 + 7;
int N, M, u, v;
int x;
int arr[SIZE];
int minDist[SIZE];
int hasrailpath[SIZE][SIZE];
int conn[SIZE][SIZE];
bool vis[SIZE];
bool contains(int N) {
  if (N < 0) N = -N;
  while (N) {
    if (N % 10 == 8) return 1;
    N /= 10;
  }
  return 0;
}
bool comp(const pair<int, int>& p1, const pair<int, int>& p2) {
  if (p1.second != p2.second) return p1.second < p2.second;
  return p1.first < p2.first;
}
int main() {
  long long int a, b, c, tb, tc;
  vector<pair<int, int> > orders;
  scanf("%d", &N);
  while (N--) scanf("%d%d", &u, &v), orders.push_back(make_pair(u, v));
  sort(orders.begin(), orders.end(), comp);
  int ans = 1;
  int prev = orders[0].second;
  int sz = orders.size();
  for (int i = 1; i < sz; i++) {
    if (orders[i].first > prev) {
      ans++, prev = orders[i].second;
    }
  }
  printf("%d\n", ans);
  return 0;
}
