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;
int64_t modex(int64_t a, int64_t b, int64_t p) {
  int64_t res = 1;
  a %= p;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return res;
}
pair<int64_t, int64_t> a[500005];
int dp[500005];
bool cmp(const pair<int64_t, int64_t>& a, const pair<int64_t, int64_t>& b) {
  if (a.second == b.second)
    return a.first < b.first;
  else
    return a.second < b.second;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
  }
  sort(a, a + n, cmp);
  dp[0] = 1;
  for (int i = 1; i < n; i++) {
    int l = -1, h = i - 1, id = 0;
    while (l <= h) {
      int mid = l + (h - l) / 2;
      if (mid == -1) l = mid + 1;
      if (a[i].first >= a[mid].second + 1)
        l = mid + 1, id = mid;
      else
        h = mid - 1;
    }
    l = h - n + 1;
    if (id == -1)
      dp[i] = max(dp[i - 1], 1);
    else {
      dp[i] = max(dp[i - 1], dp[id] + 1);
    }
  }
  cout << *max_element(dp, dp + n);
  return 0;
}
