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 {
	static PrintWriter out = new PrintWriter(System.out);

	
	public static void main(String[] args) {
		FS in = new FS();
		int N = in.nextInt();
		int ar[] = new int[N];
		for(int i = 0; i < N; i++) ar[i] = in.nextInt();
		Integer[] order = new Integer[N];
		for(int i = 0; i < N; i++) order[i] = i;
	
		Arrays.sort(order, (a,b) -> (ar[a] == ar[b] ? b-a : ar[a]-ar[b]));
		long ids[] = new long[N];
		for(int i = 0; i < N; i++) ids[i] = order[i];
//		System.out.println(Arrays.toString(ids));
		
		Wave wt = new Wave(ids);
		int M = in.nextInt();
		for(int i = 0; i < M; i++) {
			int sz = in.nextInt();
			int pos = in.nextInt();
//			System.out.println(" p="+pos+"  l="+(N-sz)+" r="+N);
			int index = wt.kthSmallest(pos, N-sz, N);
			out.println(ar[(int) ids[(int)index]]);
		}
		
		out.close();
	}
	
	
	static class Wave{
		// s for smallest, b for biggest
		long s = Long.MAX_VALUE, b = Long.MIN_VALUE; int N;
		long[] arr; int[] dex, sPtr;
		Wave sWav = null, bWav = null;
		/// CONSTRUCTOR
		Wave(long[] arr){
			this(arr.clone(), null);}
		private Wave(long[] arr, int[] dex){
			/// initialization stuff //////////
			this.arr = arr;
			if(dex == null){
				dex = new int[arr.length];
				for(int n = 0; n < dex.length; n++) dex[n] = n;}
			this.dex = dex;
			N = dex.length;
			/// find the max and min //////////
			for(int n = 0; n < N; n++){
				s = Math.min(s, arr[dex[n]]);
				b = Math.max(b, arr[dex[n]]);}
			sPtr = new int[N+1];
			for(int n = 0; n < N; n++){
				sPtr[n+1] = sPtr[n];
				if(arr[dex[n]] <= s + (b-s) / 2) sPtr[n+1]++;}
			if(s == b) return;
			/// recursive calls ///////////////
			int[] sDex = new int[sPtr[N]];
			int[] bDex = new int[N - sPtr[N]];
			for(int n = 0; n < N; n++){
				if(sPtr[n] != sPtr[n+1]) sDex[sPtr[n]] = dex[n];
				else bDex[n - sPtr[n]] = dex[n];}
			sWav = new Wave(arr, sDex);
			bWav = new Wave(arr, bDex);}
		/// returns index of kth (1-based) smallest element between [l,r)
		// (ties broken by left-to-right position)
		int kthSmallest(int k, int l, int r){
			if(s == b) return
				dex[l+k-1]; // arr[dex[l+k-1]] is the value itself
			if(sPtr[r] - sPtr[l] >= k) return
				sWav.kthSmallest(k, sPtr[l], sPtr[r]);
			return
				bWav.kthSmallest(k-sPtr[r]+sPtr[l], l-sPtr[l], r-sPtr[r]);}
		/// returns amount of ints i (ss <= i <= bb) between [l, r)
		int numValsBtwn(long ss, long bb, int l, int r){
			if(ss <= s && b <= bb) return
				r - l;
			if(bb < s || b < ss || s == b) return
				0;
			return 
				sWav.numValsBtwn(ss, bb, sPtr[l], sPtr[r])
				+ bWav.numValsBtwn(ss, bb, l-sPtr[l], r-sPtr[r]);}
		/// swaps element at (zero-based) index n-1 and index n.
		/// it will runtime error if !(1 <= n < arr.length)
		void swap(int n){
			if(dex.length == arr.length){
				arr[n-1] ^= arr[n]; arr[n] ^= arr[n-1]; arr[n-1] ^= arr[n];}
			dex[n-1] ^= dex[n]; dex[n] ^= dex[n-1]; dex[n-1] ^= dex[n];
			if(s == b) return;
			switch(sPtr[n+1] - sPtr[n-1]){
			case 2:
				sWav.swap(sPtr[n]); break;
			case 0:
				bWav.swap(n - sPtr[n]); break;
			default:
				setIndex(n-1, dex[n-1]);
				setIndex(n, dex[n]);
				sPtr[n] = (sPtr[n] == sPtr[n+1])? sPtr[n-1]: sPtr[n+1];}}
		private void setIndex(int n, int newDex){
			dex[n] = newDex;
			if(s == b) return;
			if(sPtr[n+1] != sPtr[n]) sWav.setIndex(sPtr[n], newDex);
			else bWav.setIndex(n-sPtr[n], newDex);}
	}
	
	static class FS{
		BufferedReader br;
		StringTokenizer st;
		public FS() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while(st == null || !st.hasMoreElements()) {
				try {st = new StringTokenizer(br.readLine());}
				catch(Exception e) { throw null;}
			}
			return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next());}
		double nextDouble() { return Double.parseDouble(next());}
		long nextLong() { return Long.parseLong(next());}
		int[] NIA(int n) {
			int r[] = new int[n];
			for(int i = 0; i < n; i++) r[i] = nextInt();
			return r;
		}
		long[] NLA(int n) {
			long r[] = new long[n];
			for(int i = 0; i < n; i++) r[i] = nextLong();
			return r;
		}
		char[][] grid(int r, int c){
			char res[][] = new char[r][c];
			for(int i = 0; i < r; i++) {
				char l[] = next().toCharArray();
				for(int j = 0; j < c; j++) {
					res[i][j] = l[j];
				}
			}
			return res;
		}
	}
	
}
