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: Accepted)

#include <cstdio>
#include <vector>
#include <map>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <set>
#include <cstring>
#include <string>
#include <queue>
using namespace std;
#define LL long long
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define pii pair < int , int >

int n,m;
pii p[100005];
priority_queue < int , vector < int > , greater < int > > taken;

int main(){

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	cin >> n >> m;
	for ( int i = 0; i < n; i++ ) {
		cin >> p[i].fi >> p[i].se;
		p[i].fi = m - p[i].fi;
	}
	sort(p,p+n);

	int ans = 0;
	int Time = 0;
	for ( int i = 0; i < n; i++ ){
		if ( Time <= p[i].fi ) {
			taken.push(p[i].se);
			ans += p[i].se;
			Time++;
		} else {
			if ( !taken.empty() && taken.top() < p[i].se ){
				ans += p[i].se;
				ans -= taken.top();
				taken.pop();
				taken.push(p[i].se);
			}
		}
	}
	cout << ans << endl;
	return 0;
}