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 JAVA):

Solution:
import java.util.*;
import java.util.Map.Entry;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.rmi.server.UnicastRemoteObject;

import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
 
public class Main{
    static long mod=1000000007;
//    static int dx[]={-1,0,1,0};
//    static int dy[]={0,-1,0,1};

    public static void main(String[] args)   throws Exception, IOException{        
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);

        int n=sc.nextInt();
        int m=sc.nextInt();
        P p[] = new P[n];
        for (int i = 0; i < p.length; i++) {
			p[i] = new P(sc.nextInt(), sc.nextInt());
		}
        
        sort(p);
        long ans=0;
        PriorityQueue<Integer> q=new PriorityQueue<Integer>();
        for (int t=0,c=1; c <= m; c++) {
        	while(t<p.length && p[t].a<=c) {
//               	db(c,p[t].a,p[t].b);
               	q.add(-p[t].b);
        		t++;
        	}
        	if(!q.isEmpty())ans+=-q.poll();
		}
       
        out.println(ans);
        out.flush();
    }



    static boolean validpos(int x,int y,int r, int c){
        return x<r && 0<=x && y<c && 0<=y;
    }

    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }
}

//class P {
//	int x, y;
//	P(int  x, int y) {
//	  this.x=x;
//	  this.y=y;
//	}
//}


//class P {
//    int id, d;
//    P(int  id, int d) {
//        this.id=id;
//        this.d=d;
//    }
//}

class P implements Comparable<P>{
    int a,b;
    P(int a, int b) {
    	this.b=b;
    	this.a=a;
    }

    public int compareTo(P p){
    	if(a==p.a) return b-p.b;
        return -p.a+a;
    }
}

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString() throws IOException
    {
        while( st==null || !st.hasMoreTokens() )
            st = new StringTokenizer(x.readLine());
        return st.nextToken();
    }
    public int nextInt() throws IOException
    {
        return Integer.parseInt(nextString());
    }
    public int[] nextIntArray(int size) throws IOException{
        int r[] = new int[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextInt(); 
        }
        return r;
    }
    public long[] nextLongArray(int size) throws IOException{
        long r[] = new long[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextLong(); 
        }
        return r;
    }
    public long nextLong() throws IOException
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}