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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>

int main(void)
{
    int n,m,a[100000],b[100000],**demand,lap[100000-1],rest[100000],ifend,i,j,ans=0;

    scanf("%d %d",&n,&m);
    for ( i = 0; i < m; i++)
    {
        scanf("%d %d",a+i,b+i);
    }

    demand=malloc(m*sizeof(int*));
    for ( i = 0; i < m; i++)
    {
        demand[i]=malloc((n-1)*sizeof(int));
    }

    for ( i = 0; i < m; i++)
    {
        for ( j = a[i]-1; j <= b[i]-2; j++)
        {
            demand[i][j]=1;
        }
    }

    for ( i = 0; i < m; i++)
    {
        rest[i]=1;
    }

    do
    {
        for ( j = 0; j < n-1; j++)
        {
            lap[j]=0;
            for ( i = 0; i < m; i++)
            {
                if(rest[i]==1&&demand[i][j]==1) lap[j]++;
            }
        }

        int max=0,maxindex;
        for ( j = 0; j < n-1; j++)
        {
            if(max<lap[j])
            {
                max=lap[j];
                maxindex=j;
            }
        }

        for ( i = 0; i < m; i++)
        {
            if(rest[i]==1&&demand[i][maxindex]==1) rest[i]=0;
        }
        
        ifend=0;
        for ( i = 0; i < m; i++)
        {
            ifend+=rest[i];
        }

        ans++;
    } while (ifend!=0);

    printf("%d",ans);

    for ( i = 0; i < m; i++)
    {
        free((void *)demand[i]);
    }
    free((void *)demand);

    return 0;
}