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: Runtime Error)

#include <stdio.h>
typedef long long ll;

int main(){
    int n,m;
    scanf("%d %d",&n,&m);
    m++;
    int a[n];
    int b[n];
    int c[n][m];
    int x,max,ans=0;;
    for(int i=0;i<n;i++){
        scanf("%d %d",&a[i],&b[i]);
        for(int j=0;j<m;j++){
            if(j<m-a[i]){
                c[i][j] = b[i];
            }else c[i][j] = 0;
        }
    }
   /* for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            printf("%d ",c[i][j]);
        }
        puts("");
    }*/
    for(int j=0;j<m;j++){
        max=-1;
        for(int i=0;i<n;i++){
//            printf("%d ",c[i][j]);
            if(max<c[i][j]){
                max=c[i][j];
                x=i;
            }
            if(max==c[i][j]){
                if(a[i]>a[x]){
                    x=i;
                }
            }
        }
        ans+=max;
//        printf("[%d]\n",x);
        for(int i=0;i<m;i++) c[x][i] = 0;
    }
    printf("%d",ans);
}