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 pw(long long int a, long long int b, long long int md) {
  if (!b) return 1;
  if (b & 1) return a * pw(a * a % md, b / 2, md) % md;
  return pw(a * a % md, b / 2, md) % md;
}
const long long int MAXN = 5e5 + 10;
const long long int INF = 8e18;
const long long int MOD = 1e9 + 7;
const long long int LOG = 32;
const long long int BLOCK = 320;
const long double PI = acos(-1);
const long double SADAT = 7.84e-17;
long long int n, dp[MAXN];
pair<long long int, long long int> a[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i].second >> a[i].first;
  }
  sort(a, a + n);
  dp[0] = 1;
  for (long long int i = 1; i < n; i++) {
    long long int ptr =
        lower_bound(a, a + i,
                    pair<long long int, long long int>(a[i].second, 0)) -
        a;
    if (a[ptr].first >= a[i].second) ptr--;
    if (ptr < 0) {
      dp[i] = dp[i - 1];
      continue;
    }
    dp[i] = max(dp[i - 1], dp[ptr] + 1);
  }
  cout << dp[n - 1] << '\n';
  return 0;
}
