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 <algorithm>
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;
// const int INF = (1 << 30) - 1;
// const ll LL_INF = (1LL << 62) - 1;
// const ll MOD = 1e9 + 7;

ll N, M;
map<ll, vector<ll>> mp;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M;
  for (ll i = 0; i < N; i++) {
    ll a, b;
    cin >> a >> b;
    mp[a].push_back(b);
  }

  ll ans = 0;
  priority_queue<ll> pq;
  for (ll i = 1; i <= M; i++) {
    if (!mp[i].empty()) {
      for (auto e : mp[i]) {
        pq.push(e);
      }
    }

    if (!pq.empty()) {
      ll v = pq.top();
      pq.pop();
      ans += v;
    }
  }

  cout << ans << endl;
  return 0;
}