链接：https://codeforces.ml/contest/1140/problem/C

You have a playlist consisting of nn songs. The ii-th song is characterized by two numbers titi and bibi — 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 33 songs having lengths [5,7,4][5,7,4] and beauty values [11,14,6][11,14,6] is equal to (5+7+4)⋅6=96(5+7+4)⋅6=96.

You need to choose at most kk 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 nn and kk (1≤k≤n≤3⋅1051≤k≤n≤3⋅105) – the number of songs in the playlist and the maximum number of songs you can choose, respectively.

Each of the next nn lines contains two integers titi and bibi (1≤ti,bi≤1061≤ti,bi≤106) — the length and beauty of ii-th song.

Output

Print one integer — the maximum pleasure you can get.

Examples

input

Copy

4 3
4 7
15 1
3 6
6 8

output

Copy

78

input

Copy

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

output

Copy

10000

Note

In the first test case we can choose songs 1,3,41,3,4, so the total pleasure is (4+3+6)⋅6=78(4+3+6)⋅6=78.

In the second test case we can choose song 33. The total pleasure will be equal to 100⋅100=10000100⋅100=10000.

代码：

#include<bits/stdc++.h>
using namespace std;
long long n,k,max1,p;
struct node{
	long long t,b;
}a[300001];
bool cmp(node p,node q)
{
	if(p.b!=q.b)
	return p.b<q.b;
	else
	return p.t>q.t;
}
priority_queue<long long,vector<long long>, greater<long long> > q; 
long long sum[300001];
int main()
{
	cin>>n>>k;
	for(long long i=1;i<=n;i++)
	{
		cin>>a[i].t>>a[i].b;
	}
	sort(a+1,a+1+n,cmp);
	max1=a[n].t*a[n].b;
	long long s=0;
	for(long long i=n-1;i>=1;i--)
	{
		long long r=k;
		sum[i]=a[i].t;
		q.push(a[i+1].t);
		s+=a[i+1].t;
		if(q.size()<k)
		{
			sum[i]+=s;
		}
		else
		{
			while(q.size()>=k)
			{
				p=q.top();
				s-=p;
				q.pop();
			}
			sum[i]+=s;
		}
		max1=max(max1,a[i].b*sum[i]);
	}
	cout<<max1<<endl;
}