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: Runtime Error)

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

struct data{
	int v,c;
};

bool cmp(data x,data y){
	if(x.c <= y.c)	return true;
	else			return false;
}

main(){
	ios::sync_with_stdio(false);
	cin.tie();
	
	int n,m,k;
	long long ans;
	data a[100010];
	priority_queue<int> pq;
	ans = 0;
	k = 1;
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		scanf("%d %d", &a[i].c, &a[i].v);
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=m;i++){
		while(a[k].c == i){
			pq.push(a[k].v);	
			k++;
		}
		if(!pq.empty()){
			ans += pq.top();
			pq.pop();
		}
	}
	cout<<ans<<'\n';
}