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)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
#define rep(i, n) for(ll i = 0;i < n;i++)
#define all(i) i.begin(), i.end()
template<class T, class U> bool cmax(T& a, U b) { if (a<b) {a = b; return true;} else return false; }
template<class T, class U> bool cmin(T& a, U b) { if (a>b) {a = b; return true;} else return false; }

int n, m;

bool comp(const pii &l, const pii &r){
  return l.second > r.second;
}

int main() {
  cin.tie(0); ios::sync_with_stdio(false);

  cin >> n >> m;
  vector<pair<ll, ll>> a(n);
  rep(i, n) {
    cin >> a[i].first >> a[i].second;
  }

  sort(all(a), comp);

//  cerr << endl;
//  rep(i, n) {
//    cerr << a[i].first << " " << a[i].second << endl;
//  }
//  cerr << endl;

  multimap<int, ll> d;
  int i = 0;
  int j = 0;
  ll ans = 0;
  while(j < n && i <= m) {
    //cerr << i << " " << j << endl;
    if (m >= i+a[j].first) {
      ans += a[j].second;
      d.insert({m-a[i].first, i});
      i++;
      j++;
    } else {
      auto ite = d.lower_bound(i);
      if (ite != d.end() && (*ite).second <= m-a[j].first) {
        auto tmp = *ite;
        d.erase(ite);
        d.insert({tmp.first, i});
        ans += a[j].second;
        i++;
        j++;
      } else {
        j++;
      }
    }
  }

  cout << ans << endl;
}