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>

struct youbou{
    int a;
    int b;
    int already;
};

int main(int argc, const char * argv[]) {
    long N = 0;
    long M = 0;
    scanf("%ld%ld", &N, &M);
    
    struct youbou you[100000];
    int hashi[100000];
    for(int i = 0; i < M; i++){
        scanf("%d%d", &you[i].a, &you[i].b);
        you[i].already = 0;
    }
    for(int i = 0; i < 100000; i++){
        hashi[i]= 0;
    }
    int count = 0;
    while(1){
        int break_flag = 1;
        for(int i = 0; i <= N; i++){
            hashi[i] = 0;
        }
        for(int i = 0; i < M; i++){
            if(you[i].already == 0){
                break_flag = 0;
            }else{
                continue;
            }
            for(int j = you[i].a; j < you[i].b; j++){
                hashi[j]++;
            }
        }
        if(break_flag == 1){
            break;
        }
        int index = 0;
        int max = 0;
        for(int i = 1; i <= N; i++){
            if(max < hashi[i]){
                index = i;
                max = hashi[i];
            }
        }
        count++;
        for(int i = 0; i < M; i++){
            if(you[i].a <= index && index < you[i].b){
                you[i].already = 1;
            }
        }
        /*
        for(int i = 0; i < M; i++){
            printf("%d ",you[i].already);
        }
        printf("\n");
         */
    }
    printf("%d\n", count);
    return 0;
}