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;
#define ALL(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000
#define mod 1000000007
typedef long long ll;
const ll LINF = 1001002003004005006ll;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int main(){
    int n,m;
    cin>>n>>m;
    vector<pair<ll,ll>> v;
    rep(i,n){
        ll a,b;
        cin>>a>>b;
        if(a>m) continue;
        v.push_back(make_pair(b,a));
    }
    sort(ALL(v));
    vector<bool> day(m+1,true);
    int k=v.size();
    ll ans=0;
    int l=0;
    int c=0;
    ll mostl=k-1;
    for(int i=k-1;(i>=0)&&(c<m);i--){
        mostl=min(mostl,v[i].first);
        if(day[m-v[i].second]){
            day[m-v[i].second]=false;
            ans+=v[i].first;
            c++;
        }
        else{
            ll st=mostl;
            while(st>=0&&(!day[st])){
                st--;
            }
            if(st>=0){
                day[st]=false;
                ans+=v[i].first;
                c++;
            }
            mostl=st;
        }
    }
    cout<<ans<<endl;
    return 0;
}