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)

/*  Don’t Quit
    
    When things go wrong as they sometimes will;
        When the road you’re trudging seems all uphill;
            When the funds are low, and the debts are high;
                And you want to smile, but you have to sigh;
                    When care is pressing you down a bit
                        Rest if you must, but don’t you quit.
*/
#include <bits/stdc++.h>
using namespace std;
#define IOS ios_base::sync_with_stdio(false);cin.tie(0)
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define all(a) a.begin(),a.end()
#define int long long
const int MOD=1e9+7;
const int N=3e5+10;
int a[N],b[N];
int32_t main()
{
    int n,m;
    cin>>n>>m;
    priority_queue<pair<int,int> >pq;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i]>>b[i];
        pq.push({b[i],a[i]});
    }
    int sum=0;
    int i=1;
    while(1)
    {
        if(pq.size()>0)
        {
            int k=pq.top().ss;
            int kk=pq.top().ff;
            //cout<<kk<<endl;
            if(i+k-1<=m)
            {
                sum+=kk;
                i++;

            }   
            
            pq.pop();
        }
        else
            break;
    }
    cout<<sum<<endl;
}