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 <iostream>
#include <vector>
#include <utility>
#include <algorithm>
#include <queue>
using namespace std;
typedef pair<long,long> pll;

int n, m, j;
long a, b, t, ans;
vector<pll> ab;
priority_queue<long> q;

int main(){
	cin >> n >> m;
	for (int i = 0; i < n; ++i){
		cin >> a >> b;
		ab.emplace_back(a, b);
	}
	sort(ab.begin(), ab.end(), [](pll& a, pll& b){return a.first < b.first;});
	ans = 0;
	j = 0;
	for (int i = 1; i <= m; ++i){
		while (ab[j].first <= i && j < n){
			q.push(ab[j].second);
			j++;
		}
		if (!q.empty()){
			t = q.top(); q.pop();
			ans += t;
		}
	}
	cout << ans;
}