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;
typedef pair<ll, ll> pll;
#define fi first
#define se second
#define endl '\n'

const int N = 1e5;

multiset<ll> aux;
pll days[N+5];

int main(){

    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    ll n, m; cin >>n >>m;
    for(int i = 0; i<n; i++){
        cin >>days[i].fi >>days[i].se;
    }

    sort(days, days+n);

    ll ans = 0, p = 0;
    for(int i = 1; i<=m; i++){
        while(p < n && days[p].fi <= i){
            aux.insert(-days[p++].se);
        }
        if(!aux.empty()){
            ans += abs(*aux.begin());
            aux.erase(aux.begin());
        }
    }

    cout <<ans <<endl;

    return 0;
}