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>
#define ll long long
#define pa pair<int,int>
using namespace std;
int n,m;
struct Ds{
    int a,b;
}w[100004];
priority_queue<pa,vector<pa>,greater<pa> >q;
priority_queue<pa>g;
int main(){
    scanf("%d%d",&n,&m);
    while(!g.empty())g.pop();
    for(int i=1;i<=n;i++){
        scanf("%d%d",&w[i].a,&w[i].b);
        g.push({w[i].b,w[i].a});
    }
    while(!q.empty())q.pop();
    ll ans=0;
    for(int i=1;i<=m;i++){
        if(g.empty())continue;
        pa tp=g.top();g.pop();
        if(tp.second+i-1<=m){
            q.push(tp);
            ans+=tp.first;
        }
    }
    printf("%lld\n",ans);
    return 0;
}