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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int

ll N,M;
int ans=0;
vector<ll> a(100005);
vector<ll> b(100005);
vector<int> flag(100005,0);

void dfs(ll back,int i,int count){
  //cout<<back<<" "<<i<<" "<<count<<endl;
  int hasi=0;
  for(int j=i+1;j<M;j++){
    if(back<=a.at(j)&& flag.at(j)==0){
      hasi=1;
      flag.at(j)=1;
      dfs(b.at(j),j,count+1);
    }
  }
  if(hasi==0){
    ans = max(ans,count);
  }
}

int main(){
  cin>>N>>M;
  
  
  for(int i=0;i<M;i++)
    cin>>a.at(i)>>b.at(i);
  
  for(int i=0;i<M;i++){
    if(flag.at(i)==0){
      flag.at(i)=1;
      dfs(b.at(i),i,1);
    }
  }
  cout<<ans<<endl;
}