Problem: 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.

Solution (in JAVA):

Solution:
import java.util.*;
import java.math.*;

// **** C. Playlist ****

public class C {
	static char [] in = new char [100000];
	public static void main (String [] arg) throws Throwable {
		int n = nextInt();
		int k = nextInt();
		int [] T = new int [n];
		int [] B = new int [n];
		Pair [] P = new Pair [n];
		//PairL [] PQ = new PairL [n];
		for (int i = 0; i<n; ++i) {
			T[i] = nextInt();
			B[i] = nextInt();
			P[i] = new Pair(T[i], B[i]);
			//PQ[i] = new PairL(T[i], B[i]);
		}
		Arrays.sort(P);
		PriorityQueue<PairL> Q = new PriorityQueue<PairL>();
		int COUNT = 1;
		long min_beauty = 0;
		long len = 0;
		long ans = 0;
		for (int i = 0; i<n; ++i) {
			//System.err.println("Reading " + i + " with " + P[i].t + " , " + P[i].b);
			min_beauty = P[i].b;
			len += P[i].t;
			if (Q.size() == k) {
				PairL ele = Q.poll();
				len -= ele.t;
			}
			Q.add(new PairL(P[i].t, P[i].b));
			//System.err.println("Ans after using " + i + " is " + min_beauty*len);
			long tmp = min_beauty * len;
			ans = Math.max(tmp, ans);
		}
		System.out.println(ans);
	}
	static class Pair implements Comparable<Pair>{
		int t; int b;
		public Pair(int x, int y) { t=x; b=y;}
		public int compareTo(Pair p) { return (b > p.b) ? -1 : ((t > p.t && b == p.b) ? -1 : 1);}
	}
	
	static class PairL implements Comparable<PairL> {
		int t; int b;
		public PairL(int x, int y) { t=x; b=y;}
		public int compareTo(PairL p) { return (t < p.t) ? -1 : 1;}
	}

	
	
	
	
	
	
	
	
	
	
	
	
	
	
	/************** FAST IO CODE FOLLOWS *****************/
	public static long nextLong() throws Throwable {
		long i = System.in.read();boolean neg = false;while (i < 45) i = System.in.read();if (i == 45) {neg=true;i=48;}i = i - 48;
		int j = System.in.read();while (j > 32) {i*=10;i+=j-48;j = System.in.read();}return (neg) ? -i : i;
	}
	public static int nextInt() throws Throwable {return (int)nextLong();}
	public static String next() throws Throwable {
		int i = 0; while (i < 42 && i != -1) i = System.in.read(); int cptr = 0; while (i >= 42) { in[cptr++] = (char)i; i = System.in.read();}
		return new String(in, 0,cptr);
	}
	/**** LIBRARIES ****/
	public static long gcdL(long a, long b) {while (b != 0) {long tmp = b;b = (a % b);a = tmp;}return a;}
	public static int gcd(int a, int b) {while (b != 0) {int tmp = b;b = (a % b);a = tmp;}return a;}
	public static int[] sieve(int LIM) {
		int i,count = 0;
		boolean [] b = new boolean [LIM];
		for (i = 2;i<LIM; ++i) if (!b[i]) {count++; for (int j = i<<1; j<LIM; j+=i) b[j] = true;}
		int [] primes = new int[count];
		for (i = 2,count=0;i<LIM;++i) if (!b[i]) primes[count++] = i;
		return primes;
	}
	public static int[] numPrimeFactors(int LIM) {
		int i,count = 0;
		int [] b = new int [LIM];
		for (i = 2;i<LIM; ++i) if (b[i] == 0) {count++; for (int j = i; j<LIM; j+=i) b[j]++;}
		return b;
	}
}

/* Full Problem Text:
  
You have a playlist consisting of n songs.
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) \cdot 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.

 */