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<bits/stdc++.h>
using namespace std;

#define Q int t; scanf("%d", &t); for(int q=1; q<=t; q++)
typedef long long int lli;
typedef pair<lli, lli> pi;


int main()
{
    int n, m;
    cin >> n >> m;
    priority_queue<lli>pq;
    vector<pi> v;
    for(int i=0; i<n; i++){
        lli a, b;
        cin >> a >> b;
        v.push_back({a, b});
    }
    sort(v.begin(), v.end());
    int j=0;
    lli ans = 0;
    for(int i=1; i<=m; i++){
        while(j<n && v[j].first <= i){
            pq.push(v[j].second);
            j++;
        }
        if(!pq.empty()){
            ans += pq.top();
            pq.pop();
        }
    }

    cout << ans << "\n";

    return 0;
}