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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#define min(a,b) (a)>(b)?b:a

int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    int a[m+1], b[m+1];
    int i;
    for(i=1;i<=m;i++) scanf("%d %d", &a[i], &b[i]);
    
    int a_min[n+1];
    int min;
    for(i=1;i<=m;i++) min = min(a[i], min);
    for(i=0;i<=n;i++) a_min[i] = 111111;
    for(i=1;i<=m;i++) a_min[a[i]] = min(a_min[a[i]], b[i]);
    
    int left, right;
    left = min, right = a_min[min];
    int count = 0;
    for(i=1;i<=n;i++){
        if(a_min[i]!=111111){
            if(right<=i){
                count++;
                left = i, right = a_min[i];
            } else if(left<i&&right>=a_min[i]){
                left = i, right = a_min[i];
            } else if(left<i&&right<a_min[i]){
                left = i;
            }
        }
    }
    count++;
    printf("%d", count);
    
    return 0;
}