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>

using namespace std;

priority_queue<int>q;

int n,m,ans,pos;

struct node{
	int timp,money;
}w[100001];

inline bool cmp(node x,node y){
	return x.timp<y.timp;
}

int main(){
	cin>>n>>m;
	for(register int i=1;i<=n;i++){
		cin>>w[i].timp;
		cin>>w[i].money;	
	}
	sort(w+1,w+1+n,cmp);
/**	for(int i=1;i<=n;i++){
	    cout<<w[i].timp<<" "<<w[i].money<<endl;
	}
    */
	pos=1;
	for(register int i=1;i<=m;i++){
		while(w[pos].timp<=i)q.push(w[pos++].money);
		if(!q.empty()){
		 //   cout<<"#"<<i<<" "<<q.top()<<" ";
			ans+=q.top();
			q.pop();
		}
		//cout<<endl;
	}
	cout<<ans;
}