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: Time Limit Exceeded)

#include <stdio.h>

int main(void){
    int N,M;
    scanf("%d %d",&N, &M);
    int A[N],B[N];
    for(int i=0;i<N;i++)
    scanf("%d %d", &A[i], &B[i]);

    int money=0,a=-1,count=0;
    for(int i=1;i<M+1;i++){
        int max=0;
        for(int j=0;j<N;j++){
            if(i>=A[j] && max<B[j]){
                max=B[j];
                a=j;
            }
            if(j==N-1){
                money += max;
                B[a]=0;
                //printf("moneyは%d maxは%d\n",money,max);
            }
        }
    }

    printf("%d\n",money);

}