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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(void){
	int n,m,i,j,k;
	if (scanf("%d", &n) == 1) {
	} else {
        printf("Failed to read integer.\n");
    }
    if (scanf("%d", &m) == 1) {
	} else {
        printf("Failed to read integer.\n");
    }
	int a[100000]={0};
	int b[100000]={0};
	int c[100000]={0};
	int max[100000]={0};
	int num=0;
    for(i=1;i<=n;i++){
    	if (scanf("%d", &a[i]) == 1) {
		} else {
        	printf("Failed to read integer.\n");
    	}
    	if (scanf("%d", &b[i]) == 1) {
		} else {
        	printf("Failed to read integer.\n");
    	}
    }
    int mm=0;
    m=m+1;
    for(i=0;i<n;i++){
    	m=m-1;
    	for(j=1;j<=n;j++){
   			if(a[j]<=m){
   				c[j]=b[j];
   			}else{
   				c[j]=0;
   			}
   		}
   		max[i]=c[1];
   		for(k=2;k<=n;k++){
   			if(c[k]>=max[i]){
   				max[i]=c[k];
   				num=k;
   			}
   		}
   		a[num]=100000;
   		mm+=max[i];
    }
    printf("%d",mm);
    return 0;
}