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.io.* ;

public class CF_1110 {

    static long INF ;
     public static void main(String[] args) {
        FastReader scn = new FastReader() ;
        int n = scn.nextInt() , k = scn.nextInt() ; 
        ArrayList<int[]> al = new ArrayList<>() ;
        for(int i = 0 ; i < n ;i++){
            al.add(new int[]{scn.nextInt() , scn.nextInt()});
        }
        Collections.sort(al , (int[] a, int[] b) -> (a[1] - b[1])) ;
        long min = (int)1e7 , tot = 0 , x = 0;
        long[] ans = new long[n] ;
        PriorityQueue<Long> pq = new PriorityQueue<>();
        for(int i = n-1 ; i >=0 ; i--){
            if(x < k - 1){
                x++;
                tot += al.get(i)[0] ;
                pq.add((long)al.get(i)[0]) ;
            } else {
                if(!pq.isEmpty() && pq.peek() < al.get(i)[0]){
                    long r = pq.poll() ;
                    pq.add((long)al.get(i)[0]) ;
                    tot -= r ;
                    tot += al.get(i)[0] ;
                    x++ ;
                }
                
            }
            ans[i] = tot ;
        }
        long ans1 = 0 ;
        for(int i = 0 ; i < n-1 ; i++){
            ans1 = Math.max(ans1 , (long)al.get(i)[1] * (ans[i+1] + al.get(i)[0])) ;
        }
        ans1 = Math.max(ans1 , (long)al.get(n-1)[1] * al.get(n-1)[0]) ;
        System.out.println(ans1);

     }
     
     
     
     
     
     static class FastReader 
     { 
         BufferedReader br; 
         StringTokenizer st; 
   
         public FastReader() 
         { 
             br = new BufferedReader(new
                      InputStreamReader(System.in)); 
         } 
   
         String next() 
         { 
             while (st == null || !st.hasMoreElements()) 
             { 
                 try
                 { 
                     st = new StringTokenizer(br.readLine()); 
                 } 
                 catch (IOException  e) 
                 { 
                     e.printStackTrace(); 
                 } 
             } 
             return st.nextToken(); 
         } 
   
         int nextInt() 
         { 
             return Integer.parseInt(next()); 
         } 
   
         long nextLong() 
         { 
             return Long.parseLong(next()); 
         } 
   
         double nextDouble() 
         { 
             return Double.parseDouble(next()); 
         } 
   
         String nextLine() 
         { 
             String str = ""; 
             try
             { 
                 str = br.readLine(); 
             } 
             catch (IOException e) 
             { 
                 e.printStackTrace(); 
             } 
             return str; 
         } 
     } 
     
 }