Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) D2. Optimal Subsequences (Hard Version) 数据结构 贪心

D2. Optimal Subsequences (Hard Version)

This is the harder version of the problem. In this version, 1≤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] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:

[11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);

[40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.

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

it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;

and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.

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

[10,20,20] lexicographically less than [10,21,1],

[7,99,99] is lexicographically less than [10,21,1],

[10,21,0] is lexicographically less than [10,21,1].

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

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

Input

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

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

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

The following m lines contain pairs of integers kj and posj (1≤k≤n, 1≤posj≤kj) — the requests.

Output

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

Examples

input

3

10 20 10

6

1 1

2 1

2 2

3 1

3 2

3 3

output

20

10

20

10

20

10

input

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

2

3

2

3

2

3

1

1

3

Note

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

for k=1: [20],

for k=2: [10,20],

for k=3: [10,20,10].

题意

给你n个数，定义长度为k的理想序列为当前k个数和最大的子序列，且这个子序列的字典序要最小。

然后现在给你q个询问，每次问你长度为ki的理想序列的第pos个数是什么

题解

理想序列的构成，显然是贪心的，每次放最大的字典序最小的数进去。

我们将询问离线之后，难点就变成如何求第k个数是多少，实际上这个就是典型的离线求第k大的题目。。。做法非常多，我才用的是树状数组的二分，这个复杂度是logn^2的，线段树上2分是logn的，这个我就懒得写了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+7;
int a[maxn],index[maxn],ans[maxn],sum[maxn];
int n;
int lowbit(int x){
	return x&(-x);
}

void update(int x,int val){
	while(x <= n){
		sum[x] += val;
		x += lowbit(x);
	}
}
int query(int x){
	int s=0;
	while(x>0){
		s += sum[x];
		x -= lowbit(x);
	}
	return s;
}
void solve(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		cin>>a[i];
		index[i]=0;
	}
	set<pair<int,int> >S;
	for(int i=1;i<=n;i++){
		S.insert(make_pair(-a[i],i));
	}
	int m;scanf("%d",&m);
	vector<pair<pair<int,int>,int>>Q;
	for(int i=0;i<m;i++){
		int x,y;scanf("%d%d",&x,&y);
		Q.push_back(make_pair(make_pair(x,y),i));
	}
	sort(Q.begin(),Q.end());
	int now = 0;
	for(int i=0;i<Q.size();i++){
		while(now<Q[i].first.first){
			now=now+1;
			pair<int,int> tmp = *S.begin();
			update(tmp.second,1);
			index[tmp.second]=1;
			S.erase(tmp);
		}
		int pos = Q[i].first.second;
		int l=1,r=n,Ans=n;
		while(l<=r){
			int mid=(l+r)/2;
			if(query(mid)>=pos){
				Ans=mid;
				r=mid-1;
			}else{
				l=mid+1;
			}
		}
		ans[Q[i].second]=a[Ans];
	}
	for(int i=0;i<m;i++){
		cout<<ans[i]<<endl;
	}
}
int main(){
	solve();
}