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)

  /* vishal shrivastava
     185023
     nit hamirpur
   */
  #include <bits/stdc++.h>
  using namespace std;
  #define minpq priority_queue <int, vector<int>, greater<int> >;
  #define maxpq  priority_queue <int>;
  #define F first
  #define S second
  #define mp make_pair
  #define pb push_back
  #define f(i,x,n) for(int i=x;i<n;i++)
  #define all(c) c.begin(),c.end()
  #define ll long long
  #define pll pair<ll,ll>
  #define pii pair<int,int>
  const int mod = 1e9+7, N = 1e5 + 10;
  int solve()
  {
     ll n,d;
     cin>>n>>d;
     vector<pii> a(n);
     vector<int> s;
     for(int i=1;i<=d;i++)
     {
         s.pb(i);
     }

    int co=0;
     for(int i=0;i<n;i++)
     {
         int t1,t2;
         cin>>t1>>t2;
          if(t1<=d)
          {a[i].S=t1;
            a[i].F=t2;
           co++;
          }
     }
     sort(all(a));
     reverse(all(a));
     /*for(int i=0;i<n;i++)
     {
         cout<<a[i].F<<" ";
     }*/
     ll ans=0;
     int d1=d-1;
     for(int i=0;i<co;i++)
     {    //cout<<a[i].S<<" ";
         auto in=upper_bound(all(s),a[i].S);
          int t=in-s.begin();
         if(in==s.end())
         {
             if(a[i].S==s[t-1])
                in--;
         }
         if(in!=s.end())
         {
             ans=ans+a[i].F;
             s.erase(in);
         }
     }


     cout<<ans;
  }

  int32_t main(){
    /*ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);*/
    int test=1;
    //cin>>test;
    while(test--)
    {
      solve();
    }
  }