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()
{
    int n, m;
    scanf("%d %d", &n, &m);
    int a[100005], b[100005];
    int i;
    for (i = 0; i < m; i++)
        scanf("%d %d", &a[i], &b[i]);
    int x[100005] = {}, y[100005] = {};
    for (i = 0; i < m; i++)
    {
        x[a[i]]++;
        x[b[i]]--;
    }
    int ans = 0;
    for (i = 0; i < n; i++) 
        y[i + 1] = y[i] + x[i + 1];
    int max;
    for (;;)
    {
        max = 0;
        for (i = 0; i <= n; i++)
            if (y[i] > y[max])
                max = i;
        if (y[max] == 0)
            break;
        ans++;
        for (i = 0; i < m; i++)
            if (a[i] <= max && b[i] > max)
                a[i] = b[i] = 0;
        for (i = 0; i <= n; i++)
            x[i] = y[i] = 0;
        for (i = 0; i < m; i++)
        {
            x[a[i]]++;
            x[b[i]]--;
        }
        for (i = 0; i < n; i++)
            y[i + 1] = y[i] + x[i + 1];
    }
    printf("%d\n", ans);
    return 0;
}