Omkar and Medians
Uh oh! Ray lost his array yet again! However, Omkar might be able to help because he thinks he has found the OmkArray of Ray's array. The OmkArray of an array $$$a$$$ with elements $$$a_1, a_2, \ldots, a_{2k-1}$$$, is the array $$$b$$$ with elements $$$b_1, b_2, \ldots, b_{k}$$$ such that $$$b_i$$$ is equal to the median of $$$a_1, a_2, \ldots, a_{2i-1}$$$ for all $$$i$$$. Omkar has found an array $$$b$$$ of size $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$, $$$-10^9 \leq b_i \leq 10^9$$$). Given this array $$$b$$$, Ray wants to test Omkar's claim and see if $$$b$$$ actually is an OmkArray of some array $$$a$$$. Can you help Ray?The median of a set of numbers $$$a_1, a_2, \ldots, a_{2i-1}$$$ is the number $$$c_{i}$$$ where $$$c_{1}, c_{2}, \ldots, c_{2i-1}$$$ represents $$$a_1, a_2, \ldots, a_{2i-1}$$$ sorted in nondecreasing order. 
In the second case of the first sample, the array $$$[4]$$$ will generate an OmkArray of $$$[4]$$$, as the median of the first element is $$$4$$$.In the fourth case of the first sample, the array $$$[3, 2, 5]$$$ will generate an OmkArray of $$$[3, 3]$$$, as the median of $$$3$$$ is $$$3$$$ and the median of $$$2, 3, 5$$$ is $$$3$$$.In the fifth case of the first sample, the array $$$[2, 1, 0, 3, 4, 4, 3]$$$ will generate an OmkArray of $$$[2, 1, 2, 3]$$$ as   the median of $$$2$$$ is $$$2$$$  the median of $$$0, 1, 2$$$ is $$$1$$$  the median of $$$0, 1, 2, 3, 4$$$ is $$$2$$$  and the median of $$$0, 1, 2, 3, 3, 4, 4$$$ is $$$3$$$. In the second case of the second sample, the array $$$[1, 0, 4, 3, 5, -2, -2, -2, -4, -3, -4, -1, 5]$$$ will generate an OmkArray of $$$[1, 1, 3, 1, 0, -2, -1]$$$, as   the median of $$$1$$$ is $$$1$$$  the median of $$$0, 1, 4$$$ is $$$1$$$  the median of $$$0, 1, 3, 4, 5$$$ is $$$3$$$  the median of $$$-2, -2, 0, 1, 3, 4, 5$$$ is $$$1$$$  the median of $$$-4, -2, -2, -2, 0, 1, 3, 4, 5$$$ is $$$0$$$  the median of $$$-4, -4, -3, -2, -2, -2, 0, 1, 3, 4, 5$$$ is $$$-2$$$  and the median of $$$-4, -4, -3, -2, -2, -2, -1, 0, 1, 3, 4, 5, 5$$$ is $$$-1$$$ For all cases where the answer is NO, it can be proven that it is impossible to find an array $$$a$$$ such that $$$b$$$ is the OmkArray of $$$a$$$.
Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Description of the test cases follows.The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$) — the length of the array $$$b$$$.The second line contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-10^9 \leq b_i \leq 10^9$$$) — the elements of $$$b$$$.It is guaranteed the sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$. 
For each test case, output one line containing YES if there exists an array $$$a$$$ such that $$$b_i$$$ is the median of $$$a_1, a_2, \dots, a_{2i-1}$$$ for all $$$i$$$, and NO otherwise. The case of letters in YES and NO do not matter (so yEs and No will also be accepted).
import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    arr = list(map(int, input().split()))
    (left, right) = ([-10 ** 9 - 1], [10 ** 9 + 1])
    prev = arr[0]
    for i in range(1, n):
        if arr[i] > prev:
            if arr[i] < right[-1]:
                left.append(prev)
            elif arr[i] == right[-1]:
                right.pop()
                left.append(prev)
            else:
                return 'NO'
        elif arr[i] < prev:
            if arr[i] > left[-1]:
                right.append(prev)
            elif arr[i] == left[-1]:
                left.pop()
                right.append(prev)
            else:
                return 'NO'
        prev = arr[i]
    return 'YES'
for _ in range(int(input())):
    print(solve())
for s in [*open(0)][2::2]:
    (m, *a) = map(int, s.split())
    l = [-1000000000.0]
    r = [1000000000.0]
    o = 0
    for x in a:
        if x > m:
            if x > r[-1]:
                o = 1
                break
            l += (m,)
            if x == r[-1]:
                r.pop()
        elif x < m:
            if x < l[-1]:
                o = 1
                break
            r += (m,)
            if x == l[-1]:
                l.pop()
        m = x
    print('yNeOs'[o::2])
from sys import stdin, stdout
import collections, heapq, bisect, math
input = stdin.readline

def rint():
    return int(input())

def rstr():
    return input().strip()

def rlstr():
    return list(input().strip().split())

def rlint():
    return list(map(int, input().split()))

def main():
    n = rint()
    b = rlint()
    (left, right) = ([-float('inf')], [float('inf')])
    for i in range(1, n):
        prev = b[i - 1]
        if prev != b[i]:
            if b[i] > prev:
                if b[i] > right[-1]:
                    print('no')
                    return
                left.append(prev)
                if b[i] == right[-1]:
                    right.pop()
            else:
                if b[i] < left[-1]:
                    print('no')
                    return
                right.append(prev)
                if b[i] == left[-1]:
                    left.pop()
    print('yes')
if __name__ == '__main__':
    for _ in range(rint()):
        main()