CodeForces413B

B. T-shirt buying

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A new pack of n t-shirts came to a shop. Each of the t-shirts is characterized by three integers p_i, a_i and b_i, where p_i is the price of the i-th t-shirt, a_i is front color of the i-th t-shirt and b_i is back color of the i-th t-shirt. All values p_i are distinct, and values a_i and b_i are integers from 1 to 3.

m buyers will come to the shop. Each of them wants to buy exactly one t-shirt. For the j-th buyer we know his favorite color c_j.

A buyer agrees to buy a t-shirt, if at least one side (front or back) is painted in his favorite color. Among all t-shirts that have colors acceptable to this buyer he will choose the cheapest one. If there are no such t-shirts, the buyer won't buy anything. Assume that the buyers come one by one, and each buyer is served only after the previous one is served.

You are to compute the prices each buyer will pay for t-shirts.

Input

The first line contains single integer n (1 ≤ n ≤ 200 000) — the number of t-shirts.

The following line contains sequence of integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ 1 000 000 000), where p_i equals to the price of the i-th t-shirt.

The following line contains sequence of integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 3), where a_i equals to the front color of the i-th t-shirt.

The following line contains sequence of integers b₁, b₂, ..., b_n (1 ≤ b_i ≤ 3), where b_i equals to the back color of the i-th t-shirt.

The next line contains single integer m (1 ≤ m ≤ 200 000) — the number of buyers.

The following line contains sequence c₁, c₂, ..., c_m (1 ≤ c_j ≤ 3), where c_j equals to the favorite color of the j-th buyer. The buyers will come to the shop in the order they are given in the input. Each buyer is served only after the previous one is served.

Output

Print to the first line m integers — the j-th integer should be equal to the price of the t-shirt which the j-th buyer will buy. If the j-th buyer won't buy anything, print -1.

Examples

input

5
300 200 400 500 911
1 2 1 2 3
2 1 3 2 1
6
2 3 1 2 1 1

output

200 400 300 500 911 -1 

input

2
1000000000 1
1 1
1 2
2
2 1

output

1 1000000000 

有n件T恤，告诉你T恤价格p（并且价格互不相同），正面颜色a，背面颜色b（只有三种颜色）。

有m位顾客，顾客要求的颜色c只要（c==a||c==b)，在此情况下顾客会选择最便宜的一件，同时输出其价格，若没有满足条件的则输出-1.

n和m的最大值为200000，n*m直接做必然会超时。

最好的方法是将价格和颜色捆绑起来，按价格大小排序，不过对于set容器的运用不够熟练所以做不到。

注意到颜色只有1，2，3这三种，所以我们可以用三个set代表三种颜色来储存价格，也算一种”另类捆绑“了。

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,i,x,m;
    set<int> s1,s2,s3;
    set<int>::iterator it;
    int p[200005];
    cin>>n;
    for(i=0;i<n;i++)
        cin>>p[i];
    for(i=0;i<n;i++)
    {
        cin>>x;
        if(x==1)
            s1.insert(p[i]);
        else if(x==2)
            s2.insert(p[i]);
        else
            s3.insert(p[i]);
    }
    for(i=0;i<n;i++)
    {
        cin>>x;
        if(x==1)
            s1.insert(p[i]);
        else if(x==2)
            s2.insert(p[i]);
        else
            s3.insert(p[i]);
    }
    cin>>m;
    for(i=0;i<m;i++)
    {
        cin>>x;
        if(x==1)
        {
            if(s1.size()>0)
            {
                it=s1.begin();
                cout<<*it<<" ";
                s2.erase(*it);
                s3.erase(*it);
                s1.erase(it);
            }
            else
                cout<<"-1 ";
        }
        else if(x==2)
        {
           if(s2.size()>0)
            {
                it=s2.begin();
                cout<<*it<<" ";
                s1.erase(*it);
                s3.erase(*it);
                s2.erase(it);
            }
            else
                cout<<"-1 ";
        }
        else
        {
            if(s3.size()>0)
            {
                it=s3.begin();
                cout<<*it<<" ";
                s1.erase(*it);
                s2.erase(*it);
                s3.erase(it);
            }
            else
                cout<<"-1 ";
        }
    }
    return 0;
}

注意一下容器的清空顺序即可，下面附上第一种方法的代码

#include<iostream>
#include<vector>
#include<set>
#define MAX 200000
using namespace std;
int main()
{
    int n,m,i,x,c[MAX+5];
    cin>>n;
    vector<int> p(n);
    vector< set<int> > v(4);
    for(i=0;i<n;i++)
        cin>>p[i];
    for(i=0;i<n;i++)
        cin>>x,v[x].insert(p[i]);
    for(i=0;i<n;i++)
        cin>>x,v[x].insert(p[i]);
    cin>>m;
    for(i=0;i<m;i++)
        cin>>c[i];
    for(i=0;i<m;i++)
    {
        if(!v[c[i]].empty())
        {
            n=*v[c[i]].begin(),cout<<n<<" ";
            for(x=1;x<4;x++)
                v[x].erase(n);
        }
        else cout<<"-1 ";
    }
 return 0;
}