Problem:
Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

Take and complete the first job today. You will earn the reward of 3 after four days from today.

Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

Sample Input 2

5 3
1 2
1 3
1 4
2 1
2 3

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0

Solution in C (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>
int main(void){
    
    // D - Summer Vacation
    
    // 21:39
    // 21:52
    
    int n;
    int m;
    int a[999999];
    int b[999999];
    int day = 0;
    int tmpi = 0;
    int tmpb = 0;
    int ans = 0;
    int check;
    
    scanf("%d %d", &n, &m);
    
    for(int j=1; j<n+1; j++){
        
        scanf("%d %d", &a[j], &b[j]);
        
    }
    
    for(day=0; day<=m; day++){ // day loop
    
    tmpi = 0; // reset
    tmpb = 0;
        
        for(int i=1; i<=n; i++){ // task search
            
            for(int minus=0; tmpi==0; minus++){
                
                check = m - day;
                check -= minus;
                
                if(a[i]=check){ // day check
                
                    if(b[i]>tmpb){ // money check
                    
                        tmpi = i;
                        tmpb = b[i];
                        
                        printf("minus=%d\n", minus); //debug
                    
                    }
                
                }
                
            }
            
        }
    
    a[tmpi] = 999999;
    ans += tmpb;
    
    printf("day%d >> i=%d  b[i]=%d\n", day, tmpi, tmpb); // debug
        
    }
    
    printf("%d", ans);
    
    return 0;
    
}