Problem: 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.

Examples

Input

3 4
4 3
4 1
2 2


Output

5


Input

5 3
1 2
1 3
1 4
2 1
2 3


Output

10


Input

1 1
2 1


Output

0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N=1e6+5;
vector<int> ve[N];
priority_queue<int> qe;
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	int a,b;
	for(int i=1;i<=n;++i)
	{
		scanf("%d%d",&a,&b);
		if(m-a+1>0)
			ve[m-a+1].push_back(b);
	}
	int sum=0;
	for(int i=m;i>=1;--i)
	{
		for(int j=0;j<ve[i].size();++j)
		qe.push(ve[i][j]);
		if(qe.size())
		sum+=qe.top(),qe.pop();
	}
	printf("%d\n",sum);
}