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]. 
from bisect import bisect_left, bisect_right, insort_right
class SquareSkipList:
    def __init__(self, values=None, sorted_=False, square=1000, seed=42):
        inf = float("inf")
        self.square = square
        if values is None:
            self.rand_y = seed
            self.layer1 = [inf]
            self.layer0 = [[]]
        else:
            self.layer1 = layer1 = []
            self.layer0 = layer0 = []
            if not sorted_:
                values.sort()
            y = seed
            l0 = []
            for v in values:
                y ^= (y & 0x7ffff) << 13
                y ^= y >> 17
                y ^= (y & 0x7ffffff) << 5
                if y % square == 0:
                    layer0.append(l0)
                    l0 = []
                    layer1.append(v)
                else:
                    l0.append(v)
            layer1.append(inf)
            layer0.append(l0)
            self.rand_y = y

    def add(self, x):
        y = self.rand_y
        y ^= (y & 0x7ffff) << 13
        y ^= y >> 17
        y ^= (y & 0x7ffffff) << 5
        self.rand_y = y

        if y % self.square == 0:
            layer1, layer0 = self.layer1, self.layer0
            idx1 = bisect_right(layer1, x)
            layer1.insert(idx1, x)
            layer0_idx1 = layer0[idx1]
            idx0 = bisect_right(layer0_idx1, x)
            layer0.insert(idx1+1, layer0_idx1[idx0:])  # layer0 は dict で管理した方が良いかもしれない  # dict 微妙だった
            del layer0_idx1[idx0:]
        else:
            idx1 = bisect_right(self.layer1, x)
            insort_right(self.layer0[idx1], x)

    def remove(self, x):
        idx1 = bisect_left(self.layer1, x)
        layer0_idx1 = self.layer0[idx1]
        idx0 = bisect_left(layer0_idx1, x)
        if idx0 == len(layer0_idx1):
            del self.layer1[idx1]
            self.layer0[idx1] += self.layer0.pop(idx1+1)
        else:
            del layer0_idx1[idx0]

    def search_higher_equal(self, x):  # x 以上の最小の値を返す  O(log(n))
        idx1 = bisect_left(self.layer1, x)
        layer0_idx1 = self.layer0[idx1]
        idx0 = bisect_left(layer0_idx1, x)
        if idx0 == len(layer0_idx1):
            return self.layer1[idx1]
        return layer0_idx1[idx0]

    def search_higher(self, x):  # x を超える最小の値を返す  O(log(n))
        idx1 = bisect_right(self.layer1, x)
        layer0_idx1 = self.layer0[idx1]
        idx0 = bisect_right(layer0_idx1, x)
        if idx0 == len(layer0_idx1):
            return self.layer1[idx1]
        return layer0_idx1[idx0]

    def search_lower(self, x):  # x 未満の最大の値を返す  O(log(n))
        idx1 = bisect_left(self.layer1, x)
        layer0_idx1 = self.layer0[idx1]
        idx0 = bisect_left(layer0_idx1, x)
        if idx0 == 0:  # layer0_idx1 が空の場合とすべて x 以上の場合
            return self.layer1[idx1-1]
        return layer0_idx1[idx0-1]

    def pop(self, idx):
        # 小さい方から idx 番目の要素を削除してその要素を返す（0-indexed）
        # O(sqrt(n))
        # for を回すので重め、使うなら square パラメータを大きめにするべき
        layer0 = self.layer0
        s = -1
        for i, l0 in enumerate(layer0):
            s += len(l0) + 1
            if s >= idx:
                break
        if s==idx:
            layer0[i] += layer0.pop(i+1)
            return self.layer1.pop(i)
        else:
            return layer0[i].pop(idx-s)

    def __getitem__(self, item):
        layer0 = self.layer0
        s = -1
        for i, l0 in enumerate(layer0):
            s += len(l0) + 1
            if s >= item:
                break
        if s == item:
            return self.layer1[i]
        else:
            return layer0[i][item - s]

    def print(self):
        print(self.layer1)
        print(self.layer0)

import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split()))
M = int(input())
Ans = [-1] * M
KP = [list(map(int, input().split())) for _ in range(M)]
Idx_query = list(range(M))
Idx_query.sort(key=lambda x: KP[x][0])
Idx = list(range(N))
Idx.sort(key=lambda x: A[x], reverse=True)



n = 0
st = SquareSkipList()
for idx_query in Idx_query:
    k, pos = KP[idx_query]
    while n < k:
        idx = Idx[n]
        st.add(idx)
        n += 1
    #st.print()
    #print(pos)
    Ans[idx_query] = A[st[pos-1]]
    #print(A[st[pos-1]])

print("\n".join(map(str, Ans)))
