import kotlin.math.PI
import kotlin.math.atan

/**
 * Create a convex hull, we only need to do the top part. The points with the greatest angles are right next to the
 * island boundaries, so we just have check 2 corners.
 */


fun main() {
  val points = System.`in`.bufferedReader().lineSequence().drop(1)
    .map { line -> line.split(" ", limit = 2).map { it.toInt() }.let { (x, y) -> x to y } }.toList()
  val start = points.first()
  val end = points.last()
  val (east, west) = points.monotoneChainConvexUpperHull()
  println(
    (atan(west.second.toDouble() / (west.first - start.first)) / (2 * PI) * 360)
      .coerceAtLeast(atan(east.second.toDouble() / (end.first.toDouble() - east.first)) / (2 * PI) * 360)
         )
}

typealias Point = Pair<Int, Int>

fun cross(O: Point, A: Point, B: Point) =
  (A.first - O.first) * (B.second - O.second) - (A.second - O.second) * (B.first - O.first)


fun List<Point>.monotoneChainConvexUpperHull(): Pair<Point, Point> {
  if (size == 3) {
    return get(1) to get(1)
  }
  val hull = Array(2 * size) { -10000 to -10000 }
  var k = 2 // lower hull always have 2 points
  var i: Int = size - 1
  val t = k + 1
  while (i >= 0) {
    while (k >= t && cross(hull[k - 2], hull[k - 1], get(i)) <= 0) k--
    hull[k++] = get(i)
    i--
  }
  return hull[3] to hull[k - 2]
}