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;
template <class T>
inline void si(T &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c<48 | c> 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
long long bigmod(long long p, long long e, long long M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return ret;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long modinverse(long long a, long long M) { return bigmod(a, M - 2, M); }
void io() { freopen("/Users/MyMac/Desktop/in.txt", "r", stdin); }
const int N = 5e5 + 5;
vector<pair<int, int> > v;
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
  int n;
  si(n);
  for (int i = 0; i < n; i++) {
    int a, b;
    si(a);
    si(b);
    v.push_back((make_pair(a, b)));
  }
  sort((v).begin(), (v).end(), cmp);
  int prv = v[0].second;
  int o = 1;
  for (int i = 1; i < v.size(); i++) {
    if (v[i].first > prv) {
      o++;
      prv = v[i].second;
    }
  }
  printf("%d\n", o);
  return 0;
}
