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 <queue>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;
typedef pair<ll,ll> Pair;

bool compare_pair_by_first(Pair a, Pair b) {
    return a.first < b.first;
}

bool operator<(Pair &a, Pair &b) {
    return a.second < b.second;
}

int main() {
    long long n,m; cin >> n >> m;
    vector<vector<ll>> vs(100001);
    for(ll i = 0; i < n; i++) {
        ll tmpa, tmpb;
        cin >> tmpa >> tmpb;
        vs[tmpa].push_back(tmpb);
    }
    // sort(pairs.begin(), pairs.end(), compare_pair_by_first);
    priority_queue<ll> pq;
    ll current = 0;
    ll ans = 0;
    for(ll i = 1; i <= m; i++) {
        for(auto x : vs[i]) {
            pq.push(x);
        }
        if(!pq.empty()) {
            ans += pq.top();
            pq.pop();
        }
    }
    cout << ans << endl;
}