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: Wrong Answer)

#include <stdio.h>

int main(void){
	int n,m;
	int a[100005],b[100005];
	scanf("%d%d",&n,&m);
	for(int i=0; i<m; i++){
		scanf("%d%d",&a[i],&b[i]);
	}

	int count=0;
	for(int i=0; i<m; i++){
		count++;
		for(int j=0; j<m; j++){
			if(a[i]==a[j] && b[i]==b[j])continue;
			if(a[i] <= a[j] && b[j] <= b[i]){
				a[i]=0,b[i]=n+1;
				count--;
				break;
			}
			else if(a[i] < a[j] && a[j] < b[i] && b[i]<b[j]){
				b[j]=b[i];
				a[i]=0,b[i]=n+1;
				count--;
				break;
			}
			else if(a[j] < a[i] && a[i] < b[j] &&  b[j]<b[i]){
				a[j]=a[i];
				a[i]=0,b[i]=n+1;
				count--;
				break;
			}
		}
	}
	printf("%d",count);
}