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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>
#include <vector>
#include <algorithm>
#include <map>

struct startend {
  int start;
  int end;
};

int main(){
  int N, M;
  scanf("%d", &N);
  scanf("%d", &M);

  std::vector<startend > war;
  std::vector<startend > ans;
  int temp1, temp2;
  for (int i=0; i<M; i++){
    scanf("%d", &temp1);
    scanf("%d", &temp2);
    temp1--; temp2--;
    war.push_back((startend){temp1, temp2});
  }

  ans.push_back((startend){war[0].start, war[0].end});
  for (int i=1; i<M; i++){
    //printf("i=%d\n", i);
    int j = 0;
    while(j<ans.size()){
      //printf("ans[%d].start=%d\n", j, ans[j].start);
      //printf("ans[%d].end=%d\n", j, ans[j].end);
      //printf("war[%d].start=%d\n", i, war[i].start);
      //printf("war[%d].end=%d\n", i, war[i].end);
      if (ans[j].end <= war[i].start || war[i].end <= ans[j].start){
        j++; // j番目の回答候補は使えない
        //printf("not use\n");
      } else { // j番目の回答候補が使える
        //printf("can use\n");
        ans[j].start = std::max(ans[j].start, war[i].start);
        ans[j].end = std::min(ans[j].end, war[i].end);
        break;
      }
      if (j==ans.size()){
        ans.push_back((startend){war[i].start, war[i].end});
        break;
      }
    }
  }
  printf("%ld\n", ans.size());
  return 0;
}