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<utility>
#include<algorithm>
using namespace std;
typedef pair<int,int> P;
int main(){   
    int n,m;
    cin >> n >> m;
    P p[n];
    for(int i=0;i<n;i++){
        cin >> p[i].second >> p[i].first;
    }
    sort(p, p+n);
    reverse(p, p+n);
    
    int ans=0;
    int cur=0;
    for(int i=0;i<m;i++){
        while(cur<n && p[cur].second+i>m) cur++;
        if(cur>=n) break;
        if(p[cur].second+i<=m) ans+=p[cur++].first;
    }
    cout << ans << endl;
    return 0;
}