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 lli long long int
#define uli unsigned long long int
#define INF 999999999
#define rep(i,m,n) for(lli i = m;i < n;i++)
#define rrep(i,m,n) for(lli i=n-1;i > m;i--)
#define pb(n) push_back(n)
#define Sort(n) sort(n.begin(), n.end())
#define Rev(n) reverse(n.begin(),n.end())
#define RS(n) sort(n.begin(), n.end());reverse(n.begin(),n.end())
#define Out(S) cout << S << endl
#define NeOut(S) cout << S
#define HpOut(S) cout << setprecision(20) << S << endl;
#define Vecpr vector<pair<lli,lli>>
#define mod 1000000007;
int main(){
  lli N,A,B,C=0,D=0,E,M;
  cin >> N >> M;
  vector<pair<int,int>> PS(N);
  rep(i,0,N){
    cin >> A >> B;
    A=M-A;
    PS[i]=make_pair(B,-A);
  }
  Sort(PS);
  Rev(PS);
  rep(i,0,N){
    if(PS[i].second+D>0)continue;
    C+=PS[i].first;D++;
    if(D==M)break;
  }
  Out(C);
}