n = gets.to_s.to_i
a = gets.to_s.split.map(&.to_i)
b = gets.to_s.split.map(&.to_i)

temp = a.map { |val| {val, 0} }
ans = 10**9

n.times do |i|
  heap = temp.dup
  ma = 0
  
  n.times do |j|
    # Get and remove the smallest element
    min_index = heap.index(heap.min_by { |pair| pair[0] }).not_nil!
    t, u = heap.delete_at(min_index)
    
    new_val = t + b[(i + j) % n] // 2
    new_pair = {new_val, u + 1}
    heap << new_pair
    
    ma = Math.max(ma, u + 1)
  end
  
  ans = Math.min(ans, ma)
end

puts ans