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;
long long int MOD = 1e9 + 7;
long long int INF = 1e18;
long long int sub(long long int a, long long int b, long long int p = MOD) {
  return ((a % p) - (b % p) + p) % p;
}
long long int mult(long long int a, long long int b, long long int p = MOD) {
  return ((a % p) * (b % p)) % p;
}
long long int add(long long int a, long long int b, long long int p = MOD) {
  return (a % p + b % p) % p;
}
long long int fpow(long long int n, long long int k, long long int p = MOD) {
  long long int r = 1;
  while (k > 0) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
    k = k >> 1;
  }
  return r;
}
long long int inv(long long int a, long long int p = MOD) {
  return fpow(a, p - 2, p);
}
long long int fdiv(long long int a, long long int b, long long int p = MOD) {
  long long int yinv = inv(b);
  long long int ans = (a * yinv) % p;
  return ans;
}
unsigned long long int I1018 = 1000000000000000000;
const unsigned long long int I10e5 = 100000;
const long long int I10e6 = 1e6;
template <typename T>
istream &operator>>(istream &in, vector<T> &a) {
  for (auto &item : a) {
    in >> item;
  }
  return in;
}
template <typename T, typename U>
ostream &operator<<(ostream &out, pair<T, U> &a) {
  cout << a.first << " " << a.second;
  return out;
}
template <typename T, typename U>
istream &operator>>(istream &out, pair<T, U> &a) {
  cin >> a.first >> a.second;
  return out;
}
template <typename T, typename U>
ostream &operator<<(ostream &out, map<T, U> &a) {
  for (auto &item : a) {
    out << item << endl;
  }
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &a) {
  for (auto &item : a) {
    out << item << " ";
  }
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, vector<vector<T>> &a) {
  for (auto &item : a) {
    out << item << "\n";
  }
  return out;
}
std::vector<bool> is_prime;
std::vector<long long int> primes;
void sieve(long long int n) {
  is_prime.resize(n + 2, true);
  primes.clear();
  long long int p;
  for (p = 2; p * p <= n; p++) {
    if (is_prime[p]) {
      long long int i;
      for (i = p * p; i <= n; i += p) {
        is_prime[i] = false;
      }
    }
  }
  is_prime[0] = is_prime[1] = false;
  long long int i;
  for (i = 2; i <= n; i++) {
    if (is_prime[i]) {
      primes.emplace_back(i);
    }
  }
}
map<long long int, long long int> prime_factors(long long int n) {
  map<long long int, long long int> s;
  long long int i;
  long long int tc = 0;
  while (n % 2 == 0) {
    tc++;
    n /= 2;
  }
  if (tc > 0) {
    s[2] = tc;
  }
  for (i = 3; i <= sqrt(n); i += 2) {
    tc = 0;
    while (n % i == 0) {
      tc++;
      n /= i;
    }
    if (tc > 0) {
      s[i] = tc;
    }
  }
  if (n > 2) {
    s[n] += 1;
  }
  return s;
}
std::vector<long long int> fact_vec;
void fact_fun(long long int n) {
  fact_vec.resize(n + 10);
  long long int i;
  fact_vec[0] = 1;
  for (i = 1; i <= n + 2; i++) {
    fact_vec[i] = (fact_vec[i - 1] * i) % MOD;
  }
}
long long int npr(long long int n, long long int r) {
  long long int ans = fact_vec[n];
  ans /= fact_vec[n - r];
  return ans;
}
long long int ncr(long long int n, long long int r) {
  long long int ans = fact_vec[n];
  ans /= fact_vec[n - r];
  ans /= fact_vec[r];
  return ans;
}
int main() {
  long long int i, j;
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int n;
  cin >> n;
  std::vector<pair<long long int, long long int>> v(n);
  map<long long int, long long int> sm;
  for (auto &p : v) cin >> p.first >> p.second;
  std::vector<pair<long long int, long long int>> tp = v;
  sort((tp).begin(), (tp).end());
  sort((v).begin(), (v).end(), greater<pair<long long int, long long int>>());
  long long int ans = 0;
  for (auto p : v) {
    long long int s = p.first, e = p.second;
    auto it = lower_bound((tp).begin(), (tp).end(), make_pair(e, INF));
    if (it == tp.end()) {
      sm[s] = max(ans, max(1ll, sm[s]));
    } else {
      sm[s] = max(ans, max(sm[s], 1 + sm[it->first]));
    }
    ans = max(ans, sm[s]);
  }
  for (auto p : sm) {
    ans = max(ans, p.second);
  }
  cout << ans << endl;
}
