def solve(arr, queries):
  n = len(arr)
  minimums = []
  for d in queries:
    maximum = minimum = max(arr[0:d])
    for i in range(1, n - d + 1):
      # New upper bound is higher
      if arr[i + d - 1] > maximum:
        maximum = arr[i + d - 1]
      # Previous lower bound was the maximum
      elif arr[i - 1] == maximum:
        maximum = max(arr[i : i + d])
      # Else, we mantain the current maximum
      
      # We update the minimum
      if maximum < minimum:
        minimum = maximum
    minimums.append(minimum)
  return minimums