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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void) {
    int n, m;
    scanf("%d%d", &n, &m);
    int a[n], b[n];
    int sum = 0;

    for (int i = 0; i < n; i++) {
        scanf("%d%d", &a[i], &b[i]);
    }
    
    for (int i = 0; i < m; i++) {
        int max_idx = 0;
        for (int j = 0; j < n; j++) {
            if ((i + a[j] <= m) && (b[j] >= b[max_idx]) && (a[j] >= a[max_idx])) {
                max_idx = j;
            }
        }
        printf("%d\n", max_idx);
        sum += b[max_idx]; 
        b[max_idx] = 0;
    }

    printf("%d\n", sum);
    return 0;
}