K Best

Time Limit: 8000MS	Memory Limit: 65536K
Total Submissions: 10371	Accepted: 2672
Case Time Limit: 2000MS	Special Judge

Description

Demy has n jewels. Each of her jewels has some value v_i and weight w_i.

Since her husband John got broke after recent financial crises, Demy has decided to sell some jewels. She has decided that she would keep k

 best jewels for herself. She decided to keep such jewels that their specific value is as large as possible. That is, denote the specific value of some set of jewels S = {i₁, i₂, …, i_k} as

.

Demy would like to select such k jewels that their specific value is maximal possible. Help her to do so.

Input

The first line of the input file contains n — the number of jewels Demy got, and k — the number of jewels she would like to keep (1 ≤ k ≤ n ≤ 100 000).

The following n lines contain two integer numbers each — v_i and w_i (0 ≤ v_i ≤ 10⁶, 1 ≤ w_i ≤ 10⁶, both the sum of all v_i and the sum of all w_i do not exceed 10⁷

).

Output

Output k numbers — the numbers of jewels Demy must keep. If there are several solutions, output any one.

Sample Input

3 2
1 1
1 2
1 3

Sample Output

1 2

Source

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<string>  
#include<cmath>  
#include<queue>  
#include<stack>  
#include<vector>  
#include<map>  
#include<algorithm>  
using namespace std;  
#define ll long long  
#define ms(a,b)  memset(a,b,sizeof(a))  
const int M=1e5+10;  
const int inf=0x3f3f3f3f;  
const int mod=1e9+7; 
int i,j,k,n,m; 
int a[M]; 
int b[M];
double y[M];

struct node
{
	int num;
	int v;
	int w;
	double y;
	bool operator < (const node&x) const
    {
        return y>x.y;
    }
}p[M];

bool cmp(node a,node b)
{ 
	return a.y>b.y;
}
int kk[M];
bool check(double mid)
{
	for(int i=1;i<=n;i++)
		p[i].y=p[i].v-p[i].w*mid;
	sort(p+1,p+n+1);
	double sum=0;
	for(int i=1;i<=k;i++){
       sum+=p[i].y;
	   kk[i]=p[i].num;
	}
	return sum>=0;
}


int main()
{
	while(~scanf("%d%d",&n,&k)){
		if(n==0&&k==0)break;
		for(int i=1;i<=n;i++){
			scanf("%d %d",&p[i].v,&p[i].w);
		    p[i].num=i;
	     }
        double lb=0,ub=1000000000;
	    for(int i=0;i<100;i++){
		 double mid=(lb+ub)/2;
         if(check(mid))lb=mid;
		 else ub=mid;
	     }
	     for(int i=1;i<=k;i++)
			if(i>1)printf(" %d",kk[i]);
			else printf("%d",kk[i]);
		 printf("\n");
	   }
	   return 0;
}