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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.Arrays;
import java.util.Collection;
import java.io.IOException;
import java.util.Deque;
import java.io.UncheckedIOException;
import java.io.Closeable;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.ArrayDeque;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) throws Exception {
        Thread thread = new Thread(null, new TaskAdapter(), "", 1 << 27);
        thread.start();
        thread.join();
    }

    static class TaskAdapter implements Runnable {
        @Override
        public void run() {
            InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            FastInput in = new FastInput(inputStream);
            FastOutput out = new FastOutput(outputStream);
            TaskB2 solver = new TaskB2();
            solver.solve(1, in, out);
            out.close();
        }
    }

    static class TaskB2 {
        public void solve(int testNumber, FastInput in, FastOutput out) {
            int n = in.readInt();
            int[] a = new int[1 + n];
            for (int i = 1; i <= n; i++) {
                a[i] = in.readInt();
            }
            int m = in.readInt();
            Query[] qs = new Query[m];
            for (int i = 0; i < m; i++) {
                qs[i] = new Query();
                qs[i].k = in.readInt();
                qs[i].pos = in.readInt();
            }
            Query[] sortedQs = qs.clone();
            Arrays.sort(sortedQs, (x, y) -> x.k - y.k);
            Deque<Query> deque = new ArrayDeque<>(Arrays.asList(sortedQs));
            Integer[] addIndex = new Integer[n];
            for (int i = 0; i < n; i++) {
                addIndex[i] = i + 1;
            }
            Arrays.sort(addIndex, (x, y) -> a[x] == a[y] ? x - y : -(a[x] - a[y]));
            Segment seg = new Segment(1, n, a);
            for (int i = 1; i <= n; i++) {
                seg.update(addIndex[i - 1], addIndex[i - 1], 1, n);
                while (!deque.isEmpty() && deque.peekFirst().k == i) {
                    Query q = deque.removeFirst();
                    q.ans = seg.query(q.pos, 1, n);
                }
            }

            for (Query q : qs) {
                out.println(q.ans);
            }
        }

    }

    static class FastInput {
        private final InputStream is;
        private byte[] buf = new byte[1 << 13];
        private int bufLen;
        private int bufOffset;
        private int next;

        public FastInput(InputStream is) {
            this.is = is;
        }

        private int read() {
            while (bufLen == bufOffset) {
                bufOffset = 0;
                try {
                    bufLen = is.read(buf);
                } catch (IOException e) {
                    bufLen = -1;
                }
                if (bufLen == -1) {
                    return -1;
                }
            }
            return buf[bufOffset++];
        }

        public void skipBlank() {
            while (next >= 0 && next <= 32) {
                next = read();
            }
        }

        public int readInt() {
            int sign = 1;

            skipBlank();
            if (next == '+' || next == '-') {
                sign = next == '+' ? 1 : -1;
                next = read();
            }

            int val = 0;
            if (sign == 1) {
                while (next >= '0' && next <= '9') {
                    val = val * 10 + next - '0';
                    next = read();
                }
            } else {
                while (next >= '0' && next <= '9') {
                    val = val * 10 - next + '0';
                    next = read();
                }
            }

            return val;
        }

    }

    static class Query {
        int k;
        int pos;
        int ans;

    }

    static class Segment implements Cloneable {
        private Segment left;
        private Segment right;
        private int val;
        private int size;

        public void pushUp() {
            size = left.size + right.size;
        }

        public void pushDown() {
        }

        public Segment(int l, int r, int[] val) {
            if (l < r) {
                int m = (l + r) >> 1;
                left = new Segment(l, m, val);
                right = new Segment(m + 1, r, val);
                pushUp();
            } else {
                this.val = val[l];
                size = 0;
            }
        }

        private boolean covered(int ll, int rr, int l, int r) {
            return ll <= l && rr >= r;
        }

        private boolean noIntersection(int ll, int rr, int l, int r) {
            return ll > r || rr < l;
        }

        public void update(int ll, int rr, int l, int r) {
            if (noIntersection(ll, rr, l, r)) {
                return;
            }
            if (covered(ll, rr, l, r)) {
                size = 1;
                return;
            }
            pushDown();
            int m = (l + r) >> 1;
            left.update(ll, rr, l, m);
            right.update(ll, rr, m + 1, r);
            pushUp();
        }

        public int query(int k, int l, int r) {
            Segment trace = this;
            while (l < r) {
                int m = (l + r) >> 1;
                trace.pushDown();
                if (trace.left.size >= k) {
                    r = m;
                    trace = trace.left;
                } else {
                    l = m + 1;
                    k -= trace.left.size;
                    trace = trace.right;
                }
            }
            return trace.val;
        }

    }

    static class FastOutput implements AutoCloseable, Closeable {
        private StringBuilder cache = new StringBuilder(10 << 20);
        private final Writer os;

        public FastOutput(Writer os) {
            this.os = os;
        }

        public FastOutput(OutputStream os) {
            this(new OutputStreamWriter(os));
        }

        public FastOutput println(int c) {
            cache.append(c).append('\n');
            return this;
        }

        public FastOutput flush() {
            try {
                os.append(cache);
                os.flush();
                cache.setLength(0);
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
            return this;
        }

        public void close() {
            flush();
            try {
                os.close();
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
        }

        public String toString() {
            return cache.toString();
        }

    }
}

