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;

priority_queue<int> g[100010];

int main(){
	int n,m;
	cin>>n>>m;
	
	long long res=0;
	

	for (int i=0; i<n; i++){
		int a,b;
		cin>>a>>b;
		g[a].push(b);
	}
	
	for (int i=1; i<=m; i++){
		if (g[i].empty()) continue;
		res+=g[i].top();
		g[i].pop();
		
		int p=i;
		
		while ((!g[i].empty()) and p<=m){
			g[p].push(g[i].top());
			g[i].pop();
			p++;
		}
	}
	
	cout<<res<<endl;

}