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<cstdio>
#include<algorithm>
using namespace std;
struct node{
	int x,y;
}a[100010];
bool cmp(node x,node y)
{
	if(x.y==y.y) return x.x>y.x;
	return x.y>y.y;
}
int main(){
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i].x>>a[i].y;
	}
	sort(a+1,a+1+n,cmp);
	int now=1;
	long long ans=0;
	for(int i=1;i<=n;i++)
	{
		if(now>m) break;
		if(now+a[i].x-1<=m)
		{
			now++;
			ans+=a[i].y;
		}
		else continue;
	}
	cout<<ans;
	return 0;
}