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>
#define int long long
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define F first
#define S second

using namespace std;
typedef pair<int,int> P;

const int MOD=1000000007;
int INF=100100100100100;
P po[100010];
priority_queue<int> que; 
signed main(){
  int n,m;cin>>n>>m;
  rep(i,n){
    cin>>po[i].F>>po[i].S;
  }
  sort(po,po+n);
  int ima=0;
  int ans=0;
  for(int nok=1;nok<=m;nok++){
    while(ima<n && po[ima].F<=nok){
      que.push(po[ima].S);
      ima++;
    }
    if(!que.empty()){
      ans+=que.top();
      que.pop();
    }
  }
  cout<<ans<<endl;
    
  return 0;
}