C. Playlist（优先队列）

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

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 1,3,41,3,4, 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.

题意：

您有一个由n首歌曲组成的播放列表。第i首歌的特点是它的长度和美丽分别是两个数字ti和bi。听一组歌的乐趣等于这组歌的总长度乘以其中的最小美。例如，听一组长度为[5,7,4]和美丽值为[11,14,6]的三首歌的乐趣等于（5+7+4）6=96。

您最多需要从播放列表中选择k首歌曲，因此尽可能多地欣赏这些歌曲。求最大乐趣：

#include <cstdio>
#include <cstring>
#include <string>
#include <set>
#include <cmath>
#include <iostream>
#include <stack>
#include <queue>
#include <map>
#include <vector>
#include <algorithm>
//const long long mod=2147493647;
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;
#define N 300005
struct node
{
    ll x,y;
} se[N];
bool cmp(node a,node b)
{
    return a.y>b.y;
}
int main()
{
    ll n,m;
    scanf("%lld%lld",&n,&m);
    priority_queue<ll,vector<ll>,greater<ll> >q;
    ll sum=0;
    ll ans=0;
    for(ll i=0; i<n; i++)
    {
        scanf("%lld%lld",&se[i].x,&se[i].y);
    }
    sort(se,se+n,cmp);//因为乘se[i].b，所以先对其排序
    for(int i=0; i<n; i++)
    {
        q.push(se[i].x);
        sum+=se[i].x;
        if(q.size()>m)
        {//首先不能保证前m乘积个一定最大，
            sum-=q.top();
            q.pop();
        }
        ans=max(ans,sum*se[i].y);
//        printf("%lld*******\n",ans);
    }
    printf("%lld\n",ans);
    return 0;
}