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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

#define MIN(a, b) ((a) < (b) ? (a) : (b))

int n, m;
int A[100000];

int main(){
  int i;
  scanf("%d%d", &n, &m);
  for(i=0;i<m;i++){
    int a, b;
    scanf("%d%d", &a, &b);
    if(A[a-1]) A[a-1] = MIN(A[a-1], b-1);
    else A[a-1] = b-1;
  }
  int min = n;
  int j = 0;
 
  for(i=0;i<n-1;i++){
    if(A[i])  min = MIN(min, A[i]);
    if(i == min-1){
      min = n;
      j++;
    }
  }

  printf("%d\n", j);
  return 0;
}