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 <bits/stdc++.h>
#include <boost/math/common_factor.hpp>
using namespace std;
 
template< class t>
long long int gcd( t g,t h )
{
    return ( h==0 ? g : gcd(h,g%h) );
}
 
 
struct Comp {
  bool operator()(const std::pair<int, int> &a, const std::pair<int, int> &b) {
    if (a.first != b.first) {
      return a.first < b.first;
    }
    return a.second > b.second;
  }
 
}comp_functor;

 
int main() 
{
    ios_base::sync_with_stdio(false) ; cin.tie(0) ;
    
    long long int i,j,t,n,m,l,q,ans=1,f=0,x,temp,h[200005]={0},a,b,c,d;
    
    string s1,s2;
    
    vector<pair<long long,long long>>v;
    
    cin>>n>>m;
    
    for(i=0;i<m;i++)
    {
        cin>>c>>d;
        
        v.push_back(make_pair(c,d));
    }
    
    sort(v.begin(),v.end());
    
    a=v[0].first;
    b=v[0].second;
    
    for(i=1;i<m;i++)
    {
        
        c=v[i].first;
        d=v[i].second;
        
        if(b<=c)
        {
            ans++;
            a=c;
            b=d;
        }
        else if(a>=d)
        {
            ans++;
            a=c;
            b=d;
        }
        else
        {
            a=min(a,c);
            b=min(b,d);
        }
    }
    
    cout<<ans;
    
    return 0;
}