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: Accepted)

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <map> // pair
#include <functional>
using namespace std;
int N, M;
vector<pair<int, int> > pairs;
void solve(){
  sort(pairs.begin(), pairs.end());
  //for(int i = 0; i < M; i++) cout << "first = " << pairs[i].first << " second = " << pairs[i].second << endl;
  int cutPlace;
  int cnt = 0;
  for(int i = 0; i < M; i++){
    if(i == 0){
      cutPlace = pairs[i].first;
      cnt++;
    }else if(cutPlace > pairs[i].second) continue;
    else{
      cutPlace = pairs[i].first;
      cnt++;
    }
  }
  cout << cnt << endl;
  return ;
}

int main(){
  cin >> N >> M;
  pairs = vector<pair<int, int> >(M);
  for(int i = 0; i < M; i++){
    cin >> pairs[i].second >> pairs[i].first ;
  }
  solve();
  return 0;
}