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>
#define N 100010

typedef struct jobdata{
    int a;
    int b;
    struct jobdata *next;
}Job;

Job heap[N];
int top = -1;

void pushheap(Job input){
    int point,child;
    Job tmp;
    top++;
    heap[top] = input;
    point = top;
    while(point > 0){
        child = (point-1)/2;
        if(heap[child].b < heap[point].b){
            tmp = heap[point];
            heap[point] = heap[child];
            heap[child] = tmp;
        }else{
            break;
        }
        point = child;
    }
}

Job popheap(void){
    int point,cl,cr;//child left/right
    Job tmp;

    tmp = heap[top];
    heap[top] = heap[0];
    heap[0] = tmp;
    
    point = 0;
    
    while(point < top){
        cl = point*2+1;
        cr = cl+1;
        if(cl >= top){
            break;
        }
        
        if(cr >= top || heap[cl].b > heap[cr].b){
            if(heap[cl].b > heap[point].b){
                tmp = heap[cl];
                heap[cl] = heap[point];
                heap[point] = tmp;
                point = cl;
            }else{
                break;
            }
        }else{
            if(heap[cr].b > heap[point].b){
                tmp = heap[cr];
                heap[cr] = heap[point];
                heap[point] = tmp;
                point = cr;
            }else{
                break;
            }
        }
    }
    return heap[top--];
}

int main(void){
    int n,m;
    static Job data[N];
    static Job day[N] = {};
    Job *current;
    Job tmp;
    long int total = 0;
    
    int i;

    scanf("%d %d", &n, &m);
    
    for(i = 0; i < n; i++){
        scanf("%d %d", &data[i].a, &data[i].b);
        data[i].next = NULL;
        if(day[data[i].a].a == 0){
            day[data[i].a] = data[i];
        }else{
            current = &day[data[i].a];
            while(current->next){
                current = current->next;
            }
            current->next = &data[i];
        }
    }
    
    for(i = 1; i <= m; i++){
        if(day[i].a != 0){
            current = &day[i];
            while(current){
                pushheap(*current);
                current = current->next;
            }
        }
        
        if(top >= 0){
            tmp = popheap();
        
            total += tmp.b;
        }
    }
    
    printf("%ld\n", total);
    
    return 0;
}