Codeforces Round #388 (Div. 2)D. Leaving Auction(二分+思维）

D. Leaving Auction

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

There are n people taking part in auction today. The rules of auction are classical. There were n bids made, though it’s not guaranteed they were from different people. It might happen that some people made no bids at all.

Each bid is define by two integers (ai, bi), where ai is the index of the person, who made this bid and bi is its size. Bids are given in chronological order, meaning bi < bi + 1 for all i < n. Moreover, participant never makes two bids in a row (no one updates his own bid), i.e. ai ≠ ai + 1 for all i < n.

Now you are curious with the following question: who (and which bid) will win the auction if some participants were absent? Consider that if someone was absent, all his bids are just removed and no new bids are added.

Note, that if during this imaginary exclusion of some participants it happens that some of the remaining participants makes a bid twice (or more times) in a row, only first of these bids is counted. For better understanding take a look at the samples.

You have several questions in your mind, compute the answer for each of them.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 200 000) — the number of participants and bids.

Each of the following n lines contains two integers ai and bi (1 ≤ ai ≤ n, 1 ≤ bi ≤ 109, bi < bi + 1) — the number of participant who made the i-th bid and the size of this bid.

Next line contains an integer q (1 ≤ q ≤ 200 000) — the number of question you have in mind.

Each of next q lines contains an integer k (1 ≤ k ≤ n), followed by k integers lj (1 ≤ lj ≤ n) — the number of people who are not coming in this question and their indices. It is guarenteed that lj values are different for a single question.

It’s guaranteed that the sum of k over all question won’t exceed 200 000.

Output

For each question print two integer — the index of the winner and the size of the winning bid. If there is no winner (there are no remaining bids at all), print two zeroes.

Examples

Input
6
1 10
2 100
3 1000
1 10000
2 100000
3 1000000
3
1 3
2 2 3
2 1 2

Output
2 100000
1 10
3 1000

Input
3
1 10
2 100
1 1000
2
2 1 2
2 2 3

Output
0 0
1 10

Note

Consider the first sample:
• In the first question participant number 3 is absent so the sequence of bids looks as follows: 1. 1 10
2. 2 100
3. 1 10 000
4. 2 100 000
Participant number 2 wins with the bid 100 000.
• In the second question participants 2 and 3 are absent, so the sequence of bids looks: 1. 1 10
2. 1 10 000
The winner is, of course, participant number 1 but the winning bid is 10 instead of 10 000 as no one will ever increase his own bid (in this problem).
• In the third question participants 1 and 2 are absent and the sequence is: 1. 3 1 000
2. 3 1 000 000
The winner is participant 3 with the bid 1 000.
题意：有N条竞拍信息，是人的编号和竞价（保证递增），q次提问，每次排除k个人，问如果这k个人没有参加竞拍，谁会得到拍卖品，输出编号和竞拍价。无解输出0 0.
题解：记录每个人出的最大值，按val值递减排序，找出第一个未被标记的人和第二个未被标记的人，然后在第一个人的出价中找出大于第二个人出价的最小值。输出即可。不手动二分会T。
代码：

#include <bits/stdc++.h>
#define ll long long
#define babab printf("!!!\n")
using namespace std;
int n,q,k,x,y;
int maxnum,maxval,nextval;
struct node
{
    int num,val;
    bool operator < (const node &t)const{
    return val>t.val;
    }
}last[200000];
vector<int>mp[200200];
int vis[200200];
int a[200200];
void find()
{
    int ans=-1;
    int l=0;
    int r=mp[maxnum].size()-1;
    while(r-l>=0)
    {
        int mid=(r+l)>>1;
        if(mp[maxnum][mid]>=nextval)
        {
            ans=mp[maxnum][mid];
            r=mid-1;
        }
        else
            l=mid+1;
    }
    printf("%d\n",ans);
}
void solve()
{
    maxnum=maxval=nextval=-1;
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        if(vis[last[i].num]==1)continue;
        if(last[i].val==0) break;
        if(cnt==0)
        {
            cnt++;
            maxnum=last[i].num;
            maxval=last[i].val;
        }
        else
        {
            if(last[i].num==maxnum)continue;
            else
            {
                cnt++;
                nextval=last[i].val;
            }
        }
        if(cnt==2) break;
    }
    if(maxnum==-1)
    {
        printf("0 0\n");
        return ;
    }
    /*int tmp=*find_if(mp[maxnum].begin(),mp[maxnum].end(),bind2nd(greater<int>(),nextval));
    printf("%d %d\n",maxnum,tmp);*/
    printf("%d ",maxnum);
    find();
    return ;
}
int main()
{
    //freopen("in.txt","r", stdin);
    //freopen("out.txt","w", stdout);
    scanf("%d",&n);
    for(int i=1;i<=200100;i++) mp[i].clear();
    memset(last,0,sizeof(0));
    memset(vis,0,sizeof(0));
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&x,&y);
        last[x].num=x;
        last[x].val=y;
        mp[x].push_back(y);
    }
    sort(last+1,last+1+n);
    scanf("%d",&q);
    while(q--)
    {
        scanf("%d",&k);
        for(int i=1;i<=k;i++)
        {
            scanf("%d",&a[i]);
            vis[a[i]]=1;
        }
        solve();
         for(int i=1;i<=k;i++)
        {
            vis[a[i]]=0;
        }
    }
}