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 <unistd.h>

#define MAX(a, b) ((a) > (b) ? (a) : (b))

char ibuf[1400100];
char *ibufe = ibuf-1;

inline void readall(){
  int k, t = 0;
  while((k=read(STDIN_FILENO, ibuf+t, sizeof(ibuf)-t))>0) t += k;
}

int read_uint(){
  int x=0;
  while(*(++ibufe) <'0');
  do {
    x *= 10;
    x += *ibufe-'0';
  } while(*(++ibufe) >='0');

  return x;
}

char buf[6];
char *bufe = buf;

void write_uintln(int x){
  int i;
  static char tmp[6];
  if(x==0){
    *bufe++ = '0';
    *bufe++ = '\n';
    return;
  }

  for(i=0; x; i++){
    tmp[i] = '0' + x % 10;
    x /= 10;
  }
  for(i--; i >= 0; i--){
    *bufe++ = tmp[i];
  }
  *bufe++ = '\n';
}

inline void writeall(){
  int k, t = 0;
  while((k=write(STDOUT_FILENO, buf+t, bufe-buf-t))>0) t += k;
}


int n, m;
int A[100000];

int main(){
  int i;
  readall();
  n = read_uint();
  m = read_uint();
  for(i=0;i<m;i++){
    int a, b;
    a = read_uint();
    b = read_uint();
    A[a-1] = MAX(A[a-1], n-b+1);
  }
  int min = 0;
  int count = 0;
 
  for(i=0;i<n-1;i++){
    min = MAX(min, A[i]);
    if(i == n-min-1){
      min = 0;
      count++;
    }
  }

  write_uintln(count);
  writeall();
  return 0;
}