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]. 
# Binary Indexed Tree (Fenwick Tree)
class BIT():
    """一点加算、区間取得クエリをそれぞれO(logN)で答える
    add: i番目にvalを加える
    get_sum: 区間[l, r)の和を求める
    i, l, rは0-indexed
    """
    def __init__(self, n):
        self.n = n
        self.bit = [0] * (n + 1)

    def _sum(self, i):
        s = 0
        while i > 0:
            s += self.bit[i]
            i -= i & -i
        return s

    def add(self, i, val):
        """i番目にvalを加える"""
        i = i + 1
        while i <= self.n:
            self.bit[i] += val
            i += i & -i

    def get_sum(self, l, r):
        """区間[l, r)の和を求める"""
        return self._sum(r) - self._sum(l)

    def bisect_left(self, val):
        """[0, r)の和がval以下になるときの最小のrを求める"""
        sum_ = 0
        pos = 0
        bit_length = (n+1).bit_length()
        for i in range(bit_length, -1, -1):
            k = pos + (1 << i)
            if k < self.n + 1 and sum_ + self.bit[k] < val:
                sum_ += self.bit[k]
                pos += 1 << i
        return pos + 1


from operator import itemgetter


n = int(input())
a = list(map(int, input().split()))
m = int(input())
info = [list(map(int, input().split())) + [i] for i in range(m)]

b = list(zip(a, range(len(a))))
b = sorted(b, key = itemgetter(0), reverse = True)

info = sorted(info, key = itemgetter(0))
bit = BIT(n)

ans = [0]*m
cnt = 0
for i in range(m):
    k, pos, ind = info[i]
    while cnt < k:
        bit.add(b[cnt][1], 1)
        cnt += 1
    ans[ind] = a[bit.bisect_left(pos)-1]
for i in ans:
    print(i)