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: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b)
{ return *(char*)a - *(char*)b;} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b)
{ return strcmp(*(const char **)a, *(const char **)b);} /* aaa,aab.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)


int main(void)
{
	int n, m;
	scanf("%d %d", &n, &m);

	int ab[n][2];
	int i, j, k;
	for (i = 0; i < n; i++)
		scanf("%d %d", &ab[i][0], &ab[i][1]);
	
	qsort(ab, n, sizeof(*ab), acs);

	int max, ans, sly[100001] = {0};
	ans = 0;
	i = 0;
	for (j = 1; j <= m; j++) {
		for (; i < n && ab[i][0] <= j; i++)
			sly[i] = ab[i][1];
		max = 0;
		for (k = 0; k < i; k++) 
			max = max(max, sly[k]);
		ans += max;
	}

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

	return 0;
}