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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Scanner;

import javax.swing.tree.TreeNode;
class Main{


	static final int MOD = 1000000007;

	static int MAX = 10000000;
	static	ArrayList<TreeNode> tree;
	static HashSet<Integer> set;
	static int[] ans;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);			//文字の入力
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[] a = new int[n];
		int[] b = new int[n];
		Pair[] pairs = new Pair[n];
		for(int i = 0;i < n;i++){
			a[i] = sc.nextInt();
			b[i] = sc.nextInt();
			Pair pair = new Pair(a[i], b[i]);
			pairs[i] = pair;
		}
		Arrays.sort(pairs);
		long ans = 0;
		PriorityQueue<Integer> que = new PriorityQueue<>(Collections.reverseOrder());

		int index = 0;
		for(int day = 1;day <= m;day++){

			while(true){
				if(index == n || day != pairs[index].from){
					if(!que.isEmpty()){
						ans += que.poll();			//先頭
					}
//					System.out.println("day " + day + " "  +ans);
					break;
				}
				que.add(pairs[index].end);
				index++;

			}
		}

		System.out.println(ans);











	}

	static long fac[] = new long[MAX];
	static long finv[] = new long[MAX];
	static long inv[] = new long[MAX];



	//テーブルの作成
	static void COMinit() {
		fac[0] = fac[1] = 1;
		finv[0] = finv[1] = 1;
		inv[1] = 1;
		for (int i = 2; i < MAX; i++){
			fac[i] = fac[i - 1] * i % MOD;
			inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
			finv[i] = finv[i - 1] * inv[i] % MOD;
		}
	}

	// 二項係数計算
	static	long COM(int n, int k){
		if (n < k) return 0;
		if (n < 0 || k < 0) return 0;
		return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
	}








	static int upperbond(int k,ArrayList<Integer> data){		//k以下で最大の値をdataの中から探し、返す二分探索
		if(data == null){
			return -1;
		}
		int max = data.size();
		int min = -1;
		while(max-min > 1){
			int mid = (max+min)/2;
			if(data.get(mid) <= k){
				min = mid;
			}else{
				max = mid;
			}
		}


		//		if(min != -1){
		//			min = data.get(min);
		//		}

		//		System.out.println(max);
		return min;					//indexを返している
	}

	static int lowerbond(int k,ArrayList<Integer> data){		//k以上で最小の値をdataの中から探し、返す二分探索
		if(data == null){
			return -1;
		}
		//		System.out.println("lo");
		//		System.out.println("lo");
		int max = data.size();
		int min = -1;

		while(max-min > 1){
			int mid = (max+min)/2;
			if(data.get(mid) >= k){
				max = mid;
			}else{
				min = mid;
			}
		}

		return max;			//indexを返している
	}



	static int gcd(int a,int b){				//最大公約数を返す
		if(b == 0){
			return a;
		}else{
			return gcd(b, a%b);
		}
	}
	static long gcd(long a,long b){
		if(b == 0){
			return a;
		}else{
			return gcd(b, a%b);
		}
	}



	static long lcm (long a, long b) {
		long g = gcd(a,b);
		return a/g*b;
	}




}

class Pair implements Comparable{
	int from;
	int end;
	public Pair(int from,int end) {
		this.from = from;
		this.end = end;
		// TODO 自動生成されたコンストラクター・スタブ
	}

	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return  from - otherpair.from;
	}








}