Problem:
Score : 400 points

Problem Statement

There are N islands lining up from west to east, connected by N-1 bridges.

The i-th bridge connects the i-th island from the west and the (i+1)-th island from the west.

One day, disputes took place between some islands, and there were M requests from the inhabitants of the islands:

Request i: A dispute took place between the a_i-th island from the west and the b_i-th island from the west. Please make traveling between these islands with bridges impossible.

You decided to remove some bridges to meet all these M requests.

Find the minimum number of bridges that must be removed.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq a_i < b_i \leq N

All pairs (a_i, b_i) are distinct.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the minimum number of bridges that must be removed.

Sample Input 1

5 2
1 4
2 5

Sample Output 1

1

The requests can be met by removing the bridge connecting the second and third islands from the west.

Sample Input 2

9 5
1 8
2 7
3 5
4 6
7 9

Sample Output 2

2

Sample Input 3

5 10
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5

Sample Output 3

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

// -----------------------------------
// author   : MatsuTaku
// country  : Japan
// created  : 06/17/20 20:48:52
// -----------------------------------

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int n,m; cin>>n>>m;
  struct Data {
    int i;
    int a,b;
  };
  vector<Data> abr(m);
  for (int i = 0; i < m; i++) {
    abr[i].i=i;
    cin >> abr[i].a >> abr[i].b;
  }
  auto abl = abr;
  sort(abr.begin(), abr.end(), [&](auto l, auto r) {return l.b < r.b;});
  sort(abl.begin(), abl.end(), [&](auto l, auto r) {return l.a < r.a;});
  int ans = 0;
  vector<int> vis(n, false);
  int ri = m-1;
  for (int i = m-1; i >= 0; i--) {
    if (vis[abl[i].i]) continue;
    ans++;
    vis[abl[i].i] = true;
    while (ri >= 0 and abr[ri].b > abl[i].a) {
      vis[abr[ri].i] = true;
      ri--;
    }
  }
  cout << ans << endl;

  return 0;
}