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>
#pragma GCC optimize("-Ofast")
using namespace std;
long long int findgcd(long long int a, long long int b) {
  if (b == 0) return a;
  return findgcd(b, a % b);
}
bool check(long long int n, long long int s) {
  long long int tem = n;
  long long int sum = 0;
  while (tem > 0) {
    sum += tem % 10;
    tem = tem / 10;
  }
  if (abs(n - sum) >= s)
    return true;
  else
    return false;
}
bool cmp(vector<int> a, vector<int> b) {
  if (a[0] == b[0])
    return a[1] < b[1];
  else
    return a[0] < b[0];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  for (int i1 = 0; i1 < t; i1++) {
    int n;
    cin >> n;
    vector<vector<int> > arr(n, vector<int>(2)), arrn;
    for (int i = 0; i < n; i++) cin >> arr[i][0] >> arr[i][1];
    sort(arr.begin(), arr.end(), cmp);
    int mini = arr[0][1];
    int ans = 1;
    for (int i = 1; i < n; i++) {
      if (arr[i][0] > mini) {
        ans++;
        mini = arr[i][1];
      } else
        mini = min(arr[i][1], mini);
    }
    cout << ans;
  }
}
