CodeForces - 1089L 贪心

The kingdom of Lazyland is the home to nn idlers. These idlers are incredibly lazy and create many problems to their ruler, the mighty King of Lazyland.

Today kk important jobs for the kingdom (k≤nk≤n) should be performed. Every job should be done by one person and every person can do at most one job. The King allowed every idler to choose one job they wanted to do and the ii-th idler has chosen the job aiai.

Unfortunately, some jobs may not be chosen by anyone, so the King has to persuade some idlers to choose another job. The King knows that it takes bibi minutes to persuade the ii-th idler. He asked his minister of labour to calculate the minimum total time he needs to spend persuading the idlers to get all the jobs done. Can you help him?

Input

The first line of the input contains two integers nn and kk (1≤k≤n≤1051≤k≤n≤105) — the number of idlers and the number of jobs.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤k1≤ai≤k) — the jobs chosen by each idler.

The third line of the input contains nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤1091≤bi≤109) — the time the King needs to spend to persuade the ii-th idler.

Output

The only line of the output should contain one number — the minimum total time the King needs to spend persuading the idlers to get all the jobs done.

Examples

Input

8 7
1 1 3 1 5 3 7 1
5 7 4 8 1 3 5 2

Output

Input

3 3
3 1 2
5 3 4

Output

Note

In the first example the optimal plan is to persuade idlers 1, 6, and 8 to do jobs 2, 4, and 6.

In the second example each job was chosen by some idler, so there is no need to persuade anyone.

贪心思维题，把劝说时间la[i].a从小到大排序，懒汉工作种类la[i].b,第i个工作有多少懒汉想做k[i]。

最好for循环工作种类，如果第i种工作没人做，从劝说时间最少的懒汉开始扫，如果懒汉数k【i】大于1，则

劝说这个懒汉，扫完即可。

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
#define maxn 100000+100
typedef long long ll;
struct NODE
{ll a;ll b;
}la[maxn];
ll k[maxn];
bool cmp(NODE a,NODE b)
{return a.a<b.a;
}
int main()
{int n,q;scanf("%d%d",&n,&q);for(int i=1;i<=n;i++){scanf("%lld",&la[i].b);k[la[i].b]++;}for(int i=1;i<=n;i++){scanf("%lld",&la[i].a);}sort(la+1,la+1+n,cmp);
//  for(int i=1;i<=n;i++)
//  printf("%lld+++%lld\n",la[i].a,la[i].b);ll tp=1,ans=0;for(int i=1;i<=q;i++){if(k[i]==0){for(int j=tp;j<=n;j++){if(k[la[j].b]>1){//    cout<<j<<"----"<<la[j].a<<"+++"<<k[la[j].b]<<"+++"<<la[j].b<<endl;k[la[j].b]--;ans+=la[j].a;tp=j+1;break;}}}}printf("%lld\n",ans);return 0;
}

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