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 <iostream>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <utility>
#include <string>
#include <queue>
#include <array>
#include <list>
#include <cmath>
#include <algorithm>
using namespace std;

#define pii pair<int, int>
#define mp make_pair
#define pb push_back

bool func(pii a, pii b){
	if(a.second > b.second)
		return true;
	else if (a.second == b.second)
		return (a.first > b.first);
	return false;
}

int main(void){
	int n, m;
	cin >> n >> m;
	vector<pii> v;
	vector<int> x(m,0);
	for(int i = 0;i < n;i++){
		int a, b;
		cin >> a >> b;
		if(a > m) continue;
		else if(m + 1 - x[a-1] - a > 0){
			x[a-1]++;
			v.pb(mp(a,b));
		}
	}
	int len = v.size();
	sort(v.begin(), v.end(), func);
	vector<int> xxx;
	for(int i = 0;i < m;i++)
		xxx.pb(m-i);
	int ans = 0;
	int temp = INT_MAX;
	for(int i = 0,j= 0;i < len && j < m && v[i].first < temp;i++){
		if(x[v[i].first-1] > 0){
			ans+= v[i].second;
			x[v[i].first-1]--;
			xxx[v[i].first-1]--;
			j++;
			if(xxx[v[i].first-1] == 0)
				temp = v[i].first;
		}
	}
	cout << ans << endl;
}