T = gets.to_i

def is_kadomatsu(arr)
  arr.each_cons(3).each do |a, b, c|
    return false if a == b
    return false if a == c
    return false if b == c

    return false if a < b && b < c
    return false if a > b && b > c
  end

  true
end

T.times do
  n = gets.to_i
  arr = gets.split.map(&:to_i)

  idx = -1

  1.upto(n - 2) do |i|
    a = arr[i - 1]
    b = arr[i]
    c = arr[i + 1]

    if a == b || a == c || b == c
      idx = i
      break
    end

    next if a < b && b > c
    next if a > b && b < c

    idx = i
    break
  end

  ok = false
  (idx - 1..[idx + 2, n - 2].min).each do |i|
    arr[i], arr[i + 1] = arr[i + 1], arr[i]

    if is_kadomatsu(arr)
      ok = true
    end

    arr[i], arr[i + 1] = arr[i + 1], arr[i]
  end

  if ok
    puts 'Yes'
  else
    puts 'No'
  end
end