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<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<utility>
#include<set>
#include<map>
typedef long long int ll;
using namespace std;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define REP(i,n) for (int i=0;i<signed(n);i++)
#define EREP(i,n) for (int i=1;i<=(n);i++)
#define ALL(a) (a).begin(),(a).end()
#define EVEL 1
 
#ifdef EVEL
#define DEB(X) cout << #X << ":" <<X<<" " ;
#define TF(f) f ? cout<<"true  " : cout<<"false ";
#define END cout<<"\n";
#else
#define DEB(X) {}
#define TF(f) {}
#define END {}
#endif
const int MOD = 1000000007;
const ll INF = 20000000000;
typedef std::pair<int,int> P;
	ll N,M;
	P A[100010];
	bool B[10010];
	priority_queue<int> Q;
	ll ans;
int main(){
	cin>>N>>M;
    REP(i,N){
        int a,b;
        cin>>a>>b;
        A[i]={a,b};
    }
    sort(A,A+N);
    int count=1;
    REP(i,N){
        if(A[i].first!=count){
            REP(j,A[i].first-count+1){
                if(!Q.empty()){
                     ans+=Q.top();
                   Q.pop();
              }                
            }
            count=A[i].first;
        }
        Q.push(A[i].second);
        if(M<count)break;
    }
    if(count<=M){
        REP(i,M+1-count){
         if(!Q.empty()){
             ans+=Q.top();
            Q.pop();
         }
        }
    }
	cout<<ans;
}