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.*;
import java.io.*;
import java.math.BigInteger;
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
    //SOLUTION BEGIN
    //Code not meant for understanding, proceed with caution, You have been warned.
    void pre() throws Exception{}
    void solve(int TC) throws Exception{
        int n = ni(), k = ni();
        long[][] p = new long[n][];
        MyTreeSet<Long> s1 = new MyTreeSet<>(), s2 = new MyTreeSet<>();
        for(int i = 0; i< n; i++){
            p[i] = new long[]{nl(), nl()};
            s2.add(p[i][0]);
        }
        Arrays.sort(p, (long[] l1, long[] l2) -> Long.compare(l1[1], l2[1]));
        long sum = 0, ans = 0;
        for(int i = 0; i< k; i++){
            long x = s2.last();
            s2.remove(x);
            s1.add(x);
            sum+=x;
        }
        for(int i = 0; i<n; i++){
            ans = Math.max(ans, sum*p[i][1]);
            if(s2.contains(p[i][0]))s2.remove(p[i][0]);
            else if(s1.contains(p[i][0])){
                s1.remove(p[i][0]);
                sum-=p[i][0];
            }
            while(!s2.isEmpty() && s1.size()<k){
                long x = s2.last();
                s1.add(x);
                s2.remove(x);
                sum+=x;
            }
        }
        pn(ans);
    }
    class MyTreeSet<T>{
        private int size;
        private TreeMap<T, Integer> map;
        public MyTreeSet(){
            size = 0;
            map = new TreeMap<>();
        }
        public int size(){return size;}
        public int dsize(){return map.size();}
        public boolean isEmpty(){return size==0;}
        public void add(T t){
            size++;
            map.put(t, map.getOrDefault(t, 0)+1);
        }
        public boolean remove(T t){
            if(!map.containsKey(t))return false;
            size--;
            int c = map.get(t);
            if(c==1)map.remove(t);
            else map.put(t, c-1);
            return true;
        }
        public int freq(T t){return map.getOrDefault(t, 0);}
        public boolean contains(T t){return map.getOrDefault(t,0)>0;}
        public T ceiling(T t){return map.ceilingKey(t);}
        public T floor(T t){return map.floorKey(t);}
        public T first(){return map.firstKey();}
        public T last(){return map.lastKey();}
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    long mod = (long)1e9+7, IINF = (long)1e18;
    final int INF = (int)1e9, MX = (int)1e5+7;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.1415926535897932384626433832792884197169399375105820974944, eps = 1e-8;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        int T = (multipleTC)?ni():1;
        //Solution Credits: Taranpreet Singh
        pre();for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }  
            return str;
        }
    }
}