C - C.驚天浪濤殺 （CodeForces-1140C）（堆棧+貪心）

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.

 

題意：給定n個歌曲，每個歌曲有長度值len和美麗值bea。選出<=k個，使得這些歌曲中bea的最小值×len的和最大。

思路：這道題的話，既然要使bea的最小值×len的和最大，那麼我們先將bea的值從大到小排序。然後我們固定某個b[i]，則最大值爲×（），且t[j]爲所有滿足條件的t[j]中最大的k個數，然後從左到右掃描一遍，維護一個最小堆，堆裏存儲最大的k個數。每次把t[i]插入，如果堆的大小超過k就彈出堆頂，然後更新答案即可。

AC代碼：

#include <stdio.h>
#include <string>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
typedef long long ll;
const int maxx=3000010;
const int mod=10007;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
using namespace std;
struct node
{
    int len,bea;
} edge[maxx];
bool cmp(node a,node b)
{
    return a.bea>b.bea;
}
priority_queue<int,vector<int>,greater<int> >p;
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1; i<=n; i++)
        scanf("%d%d",&edge[i].len,&edge[i].bea);
    sort(edge+1,edge+n+1,cmp);
    ll ans=0,sum=0;
    for(int i=1; i<=n; i++)
    {
        p.push(edge[i].len);
        sum+=edge[i].len;
        while(p.size()>k)
        {
            sum-=p.top();
            p.pop();
        }
        ans=max(ans,sum*edge[i].bea);
    }
    printf("%lld\n",ans);
    return 0;
}