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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class b implements Runnable {
	
	int n, m;
	int[] a, ans;
	int[][] arr;
	Query[] qs;
	
	public void solve(FS in, PrintWriter out) {
		n = in.nextInt();
		a = new int[n];
		for(int i = 0; i < n; ++i) a[i] = in.nextInt();
		m = in.nextInt();
		qs = new Query[m];
		ans = new int[m];
		for(int i = 0; i < m; ++i) {
			int len = in.nextInt();
			int idx = in.nextInt() - 1;
			qs[i] = new Query(i, len, idx);
		}
		Arrays.sort(qs);
		arr = new int[n][];
		for(int i = 0; i < n; ++i) {
			arr[i] = new int[] {a[i], i};
		}
		Arrays.sort(arr, (x, y) -> {
			if(x[0] == y[0]) {
				return x[1] - y[1];
			}
			return y[0] - x[0];
		});
		int ptr = 0;
		BIT bit = new BIT(n + 10);
		for(Query cur : qs) {
			while(ptr < cur.len) {
				int idx = arr[ptr][1];
				bit.update(idx + 1, 1);
				++ptr;
			}
			int idx = cur.idx;
			int cans = bit.getKth(idx);
			ans[cur.id] = cans;
		}
		for(int i : ans) out.println(a[i]);
	}
	
	class Query implements Comparable<Query> {
		int id;
		int len, idx;
		Query(int i, int a, int b) { id = i; len = a; idx = b; }
		public int compareTo(Query o) {
			return len - o.len;
		}
	}
	
	static class BIT {
		int n;
		int[] tree;
		
		public BIT(int n) {
			this.n = n;
			tree = new int[n + 2];
		}
		
		int read(int i) {
			int sum = 0;
			while (i > 0) {
				sum += tree[i];
				i -= i & -i;
			}
			return sum;
		}
		
		void update(int i, int val) {
			while (i <= n) {
				tree[i] += val;
				i += i & -i;
			}
		}

		// if the BIT is a freq array, returns the
		// index of the kth item, or n if there are fewer
		// than k items.
		int getKth(int k) {
			int e=Integer.highestOneBit(n), o=0;
			for (; e!=0; e>>=1) {
				if (e+o<=n && tree[e+o]<=k) {
					k-=tree[e+o];
					o+=e;
				}
			}
			return o;
		}
		
	}
	
	public void run() {
		System.err.println("go!");
		FS in = new FS();
		PrintWriter out = new PrintWriter(System.out);
		solve(in, out);
		out.close();
	}
	
	public static void main(String[] args) {
		new Thread(null, new b(), "lmao", 1L<<28).start();
	}
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreElements()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}
