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>
#include <stdlib.h>

int max(int a,int b){
  if(a > b){
    return a;
  }
  return b;
}

typedef struct pair{
  int x,y;
}point;

int compare_pair(const void *a,const void *b){
  if((*(point*)a).x > (*(point*)b).x) return 1;
  if((*(point*)a).x == (*(point*)b).x)
    if((*(point*)a).y < (*(point*)b).y) return 1;
  return -1;
}


int main(void){
  int N, M;
  scanf("%d%d", &N, &M);

  point A[N];
  for (int i = 0; i < N; i++) {
    scanf("%d%d", &A[i].x, &A[i].y);
  }

  qsort(A, N, sizeof(point), compare_pair);

  int length = 0;
  int cnt = 0;
  for (int i = 1; i < N; i++) {
    if(A[i].x == A[i-1].x) cnt++;
    else{
      length = max(length, cnt);
      cnt = 0;
    }
  }
  /* printf("%d\n", length+1); */

  int arr[M][length+1];
  int data[M];
  int maximam[M];
  for (int i = 0; i < M; i++) {
    data[i] = 0;
    maximam[i] = 0;
    for (int j = 0; j < N; j++) {
      arr[i][j] = 0;
    }
  }

  cnt = 0;
  for (int i = 0; i < N; i++) {
    if(i != 0){
      if(A[i-1].x == A[i].x) cnt++;
      else cnt = 0;
    }
    arr[A[i].x-1][cnt] = A[i].y;
  }

  for (int i = 0; i < M; i++) {
    maximam[i] = arr[i][0];
  }

  int ans = maximam[0];
  data[0]++;
  maximam[0] = arr[0][data[0]];

  for (int i = 1; i < M; i++) {
    int res = 0;
    int num = 0;
    for (int j = 0; j <= i; j++) {
      if(res < maximam[j]){
        res = maximam[j];
        num = j;
      }
    }
    ans += res;
    data[num]++;
    maximam[num] = arr[num][data[num]];
  }

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

/*   printf("\n"); */
/*   for (int i = 0; i < M; i++) { */
/*     for (int j = 0; j < N; j++) { */
/*       printf("%d ", arr[i][j]); */
/*     } */
/*     printf("\n"); */
/*   } */

  return 0;
}