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;
int n,A[100010],B[100010],m;
vector<int> G[100010];
priority_queue<int> Q;
long long ans;
int main(){
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;i++){
		scanf("%d %d",&A[i],&B[i]);
		if(m>=A[i])G[m-A[i]].push_back(B[i]);
	}
	for(int i=m;i>=0;i--){
		for(int j=0;j<G[i].size();j++)Q.push(G[i][j]);
		if(!Q.empty())ans+=Q.top(),Q.pop();
	}
	cout<<ans;
	return 0;
}