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 N = 500005;
long long mod = 998244353;
long long mod2 = 1000000009;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
bool is_vowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') return 1;
  return 0;
}
double getDistance(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
long long extended_euclidean(long long a, long long b, long long &x,
                             long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long g = extended_euclidean(b, a % b, y, x);
  y -= (a / b) * x;
  return g;
}
long long power(long long base, long long p) {
  if (p == 1) return base;
  if (!p) return 1ll;
  long long ret = power(base, p / 2);
  ret *= ret;
  ret %= mod;
  if (p & 1) ret *= base;
  return ret % mod;
}
vector<pair<int, int> > v;
int dp[N], n;
int solve(int idx) {
  if (idx >= n) return 0;
  if (dp[idx] != -1) return dp[idx];
  int ret = solve(idx + 1);
  int nxt =
      (lower_bound(v.begin(), v.end(), pair<int, int>(v[idx].second + 1, 0)) -
       v.begin());
  ret = max(ret, solve(nxt) + 1);
  return dp[idx] = ret;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    v.push_back(pair<int, int>(a, b));
  }
  sort(v.begin(), v.end());
  memset(dp, -1, sizeof(dp));
  cout << solve(0);
  return 0;
}
