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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
using ll = long long;
using namespace std;

int main(){
    int n,m;cin>>n>>m;
    vector<pair<int,int>> vec(n);
    rep(i,n){
        cin>>vec[i].second>>vec[i].first;
    }
    sort(vec.begin(),vec.end(),greater<>());

    vector<bool> day(m,0);
    ll ans=0;
    int work=0;
    rep(i,n){
        int tmp=vec[i].second-1;
        while(tmp<m){
            if(day[tmp]==0){
                day[tmp]=1;
                ans+=vec[i].first;
                work++;
                break;
            }
            else{
                tmp++;
            }
        }
    }
    cout<<ans<<endl;
}