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>
using namespace std;


void sort2vectors_B_up(vector<int> &av, vector<int> &bv) {
  // sort with B in ascending order
        int n = av.size();
        vector<int> p(n), av2(n), bv2(n);
        iota(p.begin(), p.end(), 0);
        sort(p.begin(), p.end(), [&](int a, int b) { return bv[a] < bv[b]; });
        for (int i = 0; i < n; i++) {
                av2[i] = av[p[i]];
                bv2[i] = bv[p[i]];
        }
        av = av2;
        bv = bv2;
}

void printvector(vector<int> vec) {
  for_each(vec.begin(), vec.end(), [](int x) {cout << x;});
  cout << endl;
}

int main() {
  int N, M;
  cin >> N >> M;
  vector<int> A(M);
  vector<int> B(M);
  vector<int> C(N, 0), goal(N, 0);
  for (int i = 0; i < M; ++i) {
    cin >> A.at(i) >> B.at(i);
    --A.at(i);
    --B.at(i);
  }
  int count = 0;
  sort2vectors_B_up(A, B);
  int remove = 0;
  for (int i = 0; i < M; ++i) {    
    if (A.at(i) < remove) {
      continue;
    }
    remove = B.at(i);
    ++count;
  }                                        
  cout << count << endl;
}