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: Runtime Error)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define leading zero str.erase(0, min(str.find_first_not_of('0'), str.size()-1));
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
string text="abcdefghijklmnopqrstuvwxyz";
const int maxn=1e6+7;
//        .--------------.
//        | Try First One|
//        '--------------'
//                |     .--------------.
//                |     |              |
//                V     V              |
//              .--------------.       |
//              |      AC.     |<---.  |
//              '--------------'    |  |
//              (True)|  |(False)   |  |
//           .--------'  |          |  |
//           |           V          |  |
//           |  .--------------.    |  |
//           |  |   Try Again  |----'  |
//           |  '--------------'       |
//           |                         |
//           |  .--------------.       |
//           '->| Try Next One |-------'
//              '--------------'
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n,m;
    cin>>n>>m;
    vector<int>v[n+2];
    priority_queue<int>pq;
    ll ans=0;
    for(int i=0;i<n;i++){
        int a,b;
        cin>>a>>b;
        v[a].push_back(b);
    }
    for(int i=1;i<=m;i++){
        for(auto x: v[i])pq.push(x);
        if(!pq.empty()){
            ans+=pq.top();
            pq.pop();
        }
    }
    cout<<ans<<endl;
}