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 <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#include <tuple>
#define mkp make_pair
#define mkt make_tuple
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
const ll MOD=1e9+7;

int N,M;
vector<ll> A,B;

int main(){
  cin>>N>>M;
  A.resize(N);
  B.resize(N);
  rep(i,N) cin>>A[i]>>B[i];

  vector<int> ord(N,0);
  for(int i=0;i<N;i++) ord[i]=i;
  sort(ord.begin(),ord.end(),[&](int a,int b){
    if(B[a]==B[b]) return a<b;
    else return B[a]>B[b];
  });

  vector<int> num(100001,0);
  ll ans=0;
  ll rest=M;
  for(auto i:ord){
    if(rest==0) break;
    if(num[A[i]]<M-A[i]+1){
      num[A[i]]++;
      ans+=B[i];
      rest--;
    }
  }

  cout<<ans<<endl;

  return 0;
}