/**
 * Idea is to calculate the distance per color from the previous stitch and decide based on the distance to go though
 * the back or stop and start at the next stitch.
 * The timelimit seems way to high or im way of that I missed an important usecase where this won't work.
 */
fun main() {
  val input = System.`in`.bufferedReader()
  val (back, use, start) = input.readLine().split(" ").take(3).map { it.toLong() }
  val colors = input.readLine().toList()
  val pattern = input.readLine().toList()
  val startAndEnd = 2 * start

  fun List<Int>.calculateCost(): Long {
    if (isEmpty()) return 0L
    return windowed(2)
      .fold(startAndEnd + use) { acc, (left, right) ->
        acc + use + when (val diff = right - left) {
          1 -> 0
          else -> when (val backPrice = (diff - 1) * back) {
            in 0..startAndEnd -> backPrice // it is cheaper to go through the back
            else -> startAndEnd // it is cheaper to stop and start the color later
          }
        }
      }
  }

  val distance = pattern.asSequence().withIndex().groupBy { it.value }
    .mapValues { list -> list.value.map { it.index }.calculateCost() }

  colors.forEach { println(distance[it] ?: 0) } // corner case when a color is not used in the pattern
}