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)

//abc137_d.cpp
//Sat Aug 10 20:49:51 2019

#include <iostream>
#include <string>
#include <queue>
#include <map>
#include <unordered_map>
#include <vector>
#include <algorithm>
#include <math.h>
#include <set>
#define INTINF 2147483647
#define LLINF 9223372036854775807
using namespace std;
using ll=long long;
typedef pair<ll,ll> P;

bool compare_by_b(pair<int, int> a, pair<int, int> b) {
    if(a.second != b.second){
        return a.second > b.second;
    }else{
        return a.first > b.first;
    }
}

bool first_compare_by_b(pair<int, int> a, pair<int, int> b) {
    if(a.first != b.first){
        return a.first > b.first;
    }else{
        return a.second > b.second;
    }
}

int main(){
	int n,m;
	cin >> n >> m;

	vector<P> pair;
	for (int i=0;i<n;i++){
		ll a,b;
		cin >> a >> b;
		pair.push_back(make_pair(a,b));
	}

	sort(pair.begin(),pair.end(),compare_by_b);

	ll day[m+1];
	fill(day,day+m+1,0);
	int smallestday = 1;
	int flag = 0;
	for (int i=0;i<n;i++){
		for (int j=pair[i].first;j<=m;j++){
			if (day[j]==0){
				day[j] = pair[i].second;

				if (j==smallestday){
					int smallflag = 1;
					for (int k=smallestday+1;k<=m;k++){
						if (day[k]==0){
							smallestday = k;
							smallflag = 0;
						}
					}
					if (smallflag){
						flag = 1;
					}
				}

				break;
			}
		}
		if (flag){
			break;
		}
	}

	ll ans = 0;
	for (int i=1;i<=m;i++){
		ans += day[i];
//		cout << day[i] << " ";
	}

	cout << ans << endl;
}