You have a playlist consisting of n songs. The i-th song is characterized by two numbers t_i and b_i — its length and beauty respectively. The pleasure of listening to set of songs is equal to the total length of the songs in the set multiplied by the minimum beauty among them. For example, the pleasure of listening to a set of 3 songs having lengths [5, 7, 4] and beauty values [11, 14, 6] is equal to (5 + 7 + 4) ⋅ 6 = 96.

You need to choose at most k songs from your playlist, so the pleasure of listening to the set of these songs them is maximum possible.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3 ⋅ 10^5) – the number of songs in the playlist and the maximum number of songs you can choose, respectively.

Each of the next n lines contains two integers t_i and b_i (1 ≤ t_i, b_i ≤ 10^6) — the length and beauty of i-th song.

Output

Print one integer — the maximum pleasure you can get.

Examples

Input


4 3
4 7
15 1
3 6
6 8


Output


78


Input


5 3
12 31
112 4
100 100
13 55
55 50


Output


10000

Note

In the first test case we can choose songs {1, 3, 4}, so the total pleasure is (4 + 3 + 6) ⋅ 6 = 78.

In the second test case we can choose song 3. The total pleasure will be equal to 100 ⋅ 100 = 10000.
import java.util.*;

public class Main {
	static class Pair {
		int amount, nutrition;

		Pair(int amount, int nutrition) {
			this.amount = amount;
			this.nutrition = nutrition;
		}
	}

	
	public static void main(String args[]) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int k = s.nextInt();
		
		Pair array[] = new Pair[n];   // Array of Pair type
		for (int i = 0; i < n; i++)
			array[i] = new Pair(s.nextInt(), s.nextInt());
		
		// Comparator used to sort the pairs on the basis of nutrition Value in descending order
		// Starting with maximum nutrition Value
		Arrays.sort(array, (a, b) -> (b.nutrition - a.nutrition));
        
		long max = Long.MIN_VALUE, sum = 0;
		
		PriorityQueue<Integer> pq_Of_Amounts = new PriorityQueue<>();
		// Min Priority Queue to sort the amounts in increasing order(with minimum amount on top) 
		for (int i = 0; i < n; i++) {

			Pair current=array[i];     // current.nutrition has the minimum nutrition value among all pairs considered till now as they are sorted in decreasing order of nutrition
			sum += current.amount;     // Adding the amount values to sum
			pq_Of_Amounts.add(current.amount);
			
			while (pq_Of_Amounts.size() > k)     //  Since at most K foods can be selected
				sum -= pq_Of_Amounts.remove();   // Removing minimum amount values(top values) from priority Queue , subtracting from sum to maximize pleasure and ensure that at most K values selected
    
			max = Math.max(max, sum * current.nutrition);     // comparing intermediate results to get the ans 
		}

		System.out.print(max);
		s.close();
	}
}