D2. Optimal Subsequences (Hard Version) 主席树

题目链接：https://codeforces.com/contest/1262/problem/D2

将数组按大到小排序(相同大小的按下标由小到大排序)，依次将排序后的每个数在原数组中的位置放入主席树。

对于每个询问的k，pos

输出原数组中下标为query(T[0],T[k],1,len,pos)所对应的数字即可

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 200005
#define ll long long
int T[maxn*20],L[maxn*20],R[maxn*20],sum[maxn*20],tot;
ll b[maxn];
struct node{
    int pos,num;
    bool operator <(const node &w)const{
        if(num==w.num)return pos<w.pos;
        return num>w.num;
    }
}a[maxn];
inline int update(int pre,int l,int r,int x)
{
    int rt=++tot;
    L[rt]=L[pre];
    R[rt]=R[pre];
    sum[rt]=sum[pre]+1;
    if(l<r)
    {
        int mid=l+r>>1;
        if(x<=mid)L[rt]=update(L[pre],l,mid,x);
        else R[rt]=update(R[pre],mid+1,r,x);
    }
    return rt;
}
inline int query(int u,int v,int l,int r,int k)
{
    if(l>=r)return l;
    int x=sum[L[v]]-sum[L[u]],mid=l+r>>1;
    if(x>=k)return query(L[u],L[v],l,mid,k);
    else return query(R[u],R[v],mid+1,r,k-x);
}
int main()
{
    int n,m;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i].num);
        a[i].pos=i;
    }
    sort(a+1,a+1+n);
    int len=n;
    tot=0;
    for(int i=1;i<=n;i++)
    {
        b[a[i].pos]=a[i].num;
        T[i]=update(T[i-1],1,len,a[i].pos);
    }
    int k,p;
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&k,&p);
        printf("%d\n",b[query(T[0],T[k],1,len,p)]);
    }
    return 0;
}

D2k和. Optimal Subsequences (Hard Version)

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

This is the harder version of the problem. In this version, 1≤n,m≤2⋅1051≤n,m≤2⋅105. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems.

You are given a sequence of integers a=[a1,a2,…,an]a=[a1,a2,…,an] of length nn. Its subsequence is obtained by removing zero or more elements from the sequence aa (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]a=[11,20,11,33,11,20,11]:

• [11,20,11,33,11,20,11][11,20,11,33,11,20,11], [11,20,11,33,11,20][11,20,11,33,11,20], [11,11,11,11][11,11,11,11], [20][20], [33,20][33,20] are subsequences (these are just some of the long list);
• [40][40], [33,33][33,33], [33,20,20][33,20,20], [20,20,11,11][20,20,11,11] are not subsequences.

Suppose that an additional non-negative integer kk (1≤k≤n1≤k≤n) is given, then the subsequence is called optimal if:

• it has a length of kk and the sum of its elements is the maximum possible among all subsequences of length kk;
• and among all subsequences of length kk that satisfy the previous item, it is lexicographically minimal.

Recall that the sequence b=[b1,b2,…,bk]b=[b1,b2,…,bk] is lexicographically smaller than the sequence c=[c1,c2,…,ck]c=[c1,c2,…,ck] if the first element (from the left) in which they differ less in the sequence bb than in cc. Formally: there exists tt (1≤t≤k1≤t≤k) such that b1=c1b1=c1, b2=c2b2=c2, ..., bt−1=ct−1bt−1=ct−1 and at the same time bt<ctbt<ct. For example:

• [10,20,20][10,20,20] lexicographically less than [10,21,1][10,21,1],
• [7,99,99][7,99,99] is lexicographically less than [10,21,1][10,21,1],
• [10,21,0][10,21,0] is lexicographically less than [10,21,1][10,21,1].

You are given a sequence of a=[a1,a2,…,an]a=[a1,a2,…,an] and mm requests, each consisting of two numbers kjkj and posjposj (1≤k≤n1≤k≤n, 1≤posj≤kj1≤posj≤kj). For each query, print the value that is in the index posjposj of the optimal subsequence of the given sequence aa for k=kjk=kj.

For example, if n=4n=4, a=[10,20,30,20]a=[10,20,30,20], kj=2kj=2, then the optimal subsequence is [20,30][20,30] — it is the minimum lexicographically among all subsequences of length 22 with the maximum total sum of items. Thus, the answer to the request kj=2kj=2, posj=1posj=1 is the number 2020, and the answer to the request kj=2kj=2, posj=2posj=2 is the number 3030.

Input

The first line contains an integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of the sequence aa.

The second line contains elements of the sequence aa: integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109).

The third line contains an integer mm (1≤m≤2⋅1051≤m≤2⋅105) — the number of requests.

The following mm lines contain pairs of integers kjkj and posjposj (1≤k≤n1≤k≤n, 1≤posj≤kj1≤posj≤kj) — the requests.

Output

Print mm integers r1,r2,…,rmr1,r2,…,rm (1≤rj≤1091≤rj≤109) one per line: answers to the requests in the order they appear in the input. The value of rjrj should be equal to the value contained in the position posjposj of the optimal subsequence for k=kjk=kj.

Examples

input

Copy

3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3

output

Copy

20
10
20
10
20
10

input

Copy

7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4

output

Copy

2
3
2
3
2
3
1
1
3

Note

In the first example, for a=[10,20,10]a=[10,20,10] the optimal subsequences are:

• for k=1k=1: [20][20],
• for k=2k=2: [10,20][10,20],
• for k=3k=3: [10,20,10][10,20,10].