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;
typedef long long ll;
#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; i++)
#define reps(i, s, n) for(ll i = (s), i##_len = (n); i < i##_len; i++)
#define rrep(i, n) for(ll i = (n) - 1; i >= 0; i--)
#define rreps(i, e, n) for(ll i = (n) - 1; i >= (e); i--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())
#define len(x) ((ll)(x).length())
#define endl "\n"

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    // ifstream in("input.txt");
    // cin.rdbuf(in.rdbuf());
    ll n, m;
    cin >> n >> m;
    vector<pair<ll, ll>> ab(n);
    rep(i, n) cin >> ab[i].first >> ab[i].second;
    sort(all(ab), [](pair<ll, ll> &a, pair<ll, ll> &b){ return a.first < b.first; });
    auto it = ab.begin();
    priority_queue<ll> pq;
    ll ans = 0;
    rrep(i, m) {
        while((it != ab.end()) && (i <= (m - it->first))) {
            pq.push(it->second);
            it++;
        }
        if (pq.empty()) continue;
        ans += pq.top();
        pq.pop();
    }
    cout << ans << endl;
    // rep(i, n) printf("%lld:%lld\n", ab[i].first, ab[i].second);
    return 0;
}