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>
#include <stdlib.h>

int compare_int(const void *a, const void *b)
{
	return *(int*)a - *(int*)b;
}

void QSort(int *x, int *y, int left, int right);
void Swap(int *x, int *y, int i, int j);


/* クイックソートを行う */
void QSort(int *x, int *y, int left, int right)
{
	int i, j;
	int pivot;

	i = left;                      /* ソートする配列の一番小さい要素の添字 */
	j = right;                     /* ソートする配列の一番大きい要素の添字 */

	pivot = x[(left + right) / 2]; /* 基準値を配列の中央付近にとる */

	while (1) {                    /* 無限ループ */

		while (x[i] < pivot)       /* pivot より大きい値が */
			i++;                   /* 出るまで i を増加させる */

		while (pivot < x[j])       /* pivot より小さい値が */
			j--;                   /*  出るまで j を減少させる */
		if (i >= j)                /* i >= j なら */
			break;                 /* 無限ループから抜ける */

		Swap(x, y, i, j);             /* x[i] と x[j]を交換 */
		i++;                       /* 次のデータ */
		j--;
	}


	if (left < i - 1)              /* 基準値の左に 2 以上要素があれば */
		QSort(x, y, left, i - 1);     /* 左の配列を Q ソートする */
	if (j + 1 <  right)            /* 基準値の右に 2 以上要素があれば */
		QSort(x, y, j + 1, right);    /* 右の配列を Q ソートする */
}

/* 配列の要素を交換する */
void Swap(int *x, int *y, int i, int j)
{
	int temp;

	temp = x[i];
	x[i] = x[j];
	x[j] = temp;

	temp = y[i];
	y[i] = y[j];
	y[j] = temp;


}


int main(){

	int N, M, *A, *B, i, j, *W, *flag, sum=0, start, end, length, tmp;




	scanf("%d %d", &N, &M);


	A = (int *)malloc(N * sizeof(int));
	B = (int *)malloc(N * sizeof(int));

	flag = (int *)malloc(N * sizeof(int));


	for (i = 0; i < N;i++){
		scanf("%d %d", &A[i], &B[i]);
		flag[i] = 0;
	}
	QSort(B, A, 0, N - 1);



	for (i = 0; i < M; i++){
		for (j = N - 1; j >= 0; j--){
			if (i + A[j] <= M && flag[j]==0){
				flag[j] = 1;
				printf("%d %d %d\n", i, j, A[j]);
				sum += B[j];
				break;
			}
		}
	}
	printf("%d\n", sum);
	return 0;

}