This is the harder version of the problem. In this version, 1 ≤ n, m ≤ 2⋅10^5. 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=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.

For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.

Input

The first line contains an integer n (1 ≤ n ≤ 2⋅10^5) — the length of the sequence a.

The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).

The third line contains an integer m (1 ≤ m ≤ 2⋅10^5) — the number of requests.

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

Output

Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.

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]. 
import java.io.*;
import java.util.*;

public class P602D2 {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        String[] line = br.readLine().split(" ");

        Data[] a = new Data[n];
        long[] orig = new long[n];
        for(int i=0; i<n; i++) {
            a[i] = new Data(Long.parseLong(line[i]), i+1);
            orig[i] = Long.parseLong(line[i]);
        }

        int m = Integer.parseInt(br.readLine());
        Request[] reqs = new Request[m];
        for(int i=0; i<m; i++) {
            line = br.readLine().split(" ");
            reqs[i] = new Request(Integer.parseInt(line[0]), Integer.parseInt(line[1]), i);
        }

        Arrays.sort(a);
        Arrays.sort(reqs);

        FenwickTree tree = new FenwickTree(n);

        int curK = 0;
        for(int i=0; i<reqs.length; i++) {
            Request req = reqs[i];
            while(req.k > curK) {
                tree.adjust(a[curK].index, 1);
                curK++;
            }
            req.result = tree.pos(req.pos);
        }

        Arrays.sort(reqs, new IndexComparator());

        for(int i=0; i<reqs.length; i++) {
            System.out.println(orig[reqs[i].result - 1]);
        }
    }

    static class IndexComparator implements Comparator<Request> {
        @Override
        public int compare(Request a, Request b) {
            return Integer.compare(a.index, b.index);
        }
    }


    static class Request implements Comparable<Request> {
        public int k;
        public int pos;
        public int index;
        public int result = -1;
        public Request(int k, int pos, int i) {
            this.k = k;
            this.pos = pos;
            index = i;
        }

        @Override
        public int compareTo(Request o) {
            return Integer.compare(k, o.k);
        }
    }

    static class Data implements Comparable<Data> {
        public long val;
        public int index;
        public Data(long v, int i) {
            val = v;
            index = i;
        }

        // descending
        @Override
        public int compareTo(Data o) {
            if(val == o.val) {
                return Integer.compare(index, o.index);
            } else {
                return Long.compare(o.val, val);
            }
        }
    }

    static class FenwickTree {
        public int[] ft;
        FenwickTree(int n) {
            ft = new int[n+1];
        }

        public int pos(int p) {
            return pos(1, ft.length-1, p);
        }

        public int pos(int min, int max, int pos) {
            if(min == max) {
                return min;
            }
            int mid = min + (max - min) / 2;
            int sum = rsq(mid);
            if(sum >= pos) {
                return pos(min, mid, pos);
            } else {
                return pos(mid + 1, max, pos);
            }
        }

        public int rsq(int b) {
            int sum = 0;
            for(; b > 0; b-= LSOne(b))
                sum += ft[b];
            return sum;
        }

        public int rsq(int a, int b) {
            return rsq(b) - (a == 1 ? 0 : rsq(a-1));
        }

        public void adjust(int k, int v) {
            for(; k < ft.length; k += LSOne(k))
                ft[k] += v;
        }

        private int LSOne(int i) {
            return i & (-i);
        }
    }
}
