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 int long long
#define P pair<int,int>
#define PP pair<P,int>
#define F first
#define S second
#define f(i,n) for(int i=0;i<(n);i++)
#define INF LLONG_MAX/3
#define eps LDBL_EPSILON
#define mod (int)(1000000007)
#define pie 3.141592653589793238462643383279
#define Bpri priority_queue
#define Spri priority_queue<P,vector<P>,greater<P>>
#define all(x) (x).begin(),(x).end()
using namespace std;
P p[100010];
bool used[100010];
int x[100010];
signed main(){
	int ans=0;
	int a,b;
	cin>>a>>b;
	f(i,a){
		cin>>p[i].S>>p[i].F;
	}
	sort(p,p+a);
	for(int i=a-1;i>=0;i--){
		int j=p[i].S;
		if(used[j]==false){
			used[j]=true;
			x[j]=j;
			ans+=p[i].F;
			goto loop;
		}
		while(used[j]){
			if(used[j+1]==false){
				j++;
				break;
			}
			else{
				j=x[j];
			}
		}
		x[j]=x[j-1];
		x[x[j]]=j;
		loop:;
	}
	cout<<ans<<endl;
	return 0;
}