Problem:
Score : 900 points

Problem Statement

In the city of Nevermore, there are 10^8 streets and 10^8 avenues, both numbered from 0 to 10^8-1.
All streets run straight from west to east, and all avenues run straight from south to north.
The distance between neighboring streets and between neighboring avenues is exactly 100 meters.

Every street intersects every avenue. Every intersection can be described by pair (x, y), where x is avenue ID and y is street ID.

There are N fountains in the city, situated at intersections (X_i, Y_i).
Unlike normal intersections, there's a circle with radius 10 meters centered at the intersection, and there are no road parts inside this circle.

The picture below shows an example of how a part of the city with roads and fountains may look like.

City governors don't like encountering more than one fountain while moving along the same road.
Therefore, every street contains at most one fountain on it, as well as every avenue.

Citizens can move along streets, avenues and fountain perimeters.
What is the shortest distance one needs to cover in order to get from intersection (x_1, y_1) to intersection (x_2, y_2)?

Constraints

0 \leq x_1, y_1, x_2, y_2 < 10^8

1 \leq N \leq 200,000

0 \leq X_i, Y_i < 10^8

X_i \neq X_j for i \neq j

Y_i \neq Y_j for i \neq j

Intersections (x_1, y_1) and (x_2, y_2) are different and don't contain fountains.

All input values are integers.

Input

Input is given from Standard Input in the following format:

x_1 y_1 x_2 y_2
N
X_1 Y_1
X_2 Y_2
:
X_N Y_N

Output

Print the shortest possible distance one needs to cover in order to get from intersection (x_1, y_1) to intersection (x_2, y_2), in meters.
Your answer will be considered correct if its absolute or relative error doesn't exceed 10^{-11}.

Sample Input 1

1 1 6 5
3
3 2
5 3
2 4

Sample Output 1

891.415926535897938

One possible shortest path is shown on the picture below. The path starts at the blue point, finishes at the purple point and follows along the red line.

Sample Input 2

3 5 6 4
3
3 2
5 3
2 4

Sample Output 2

400.000000000000000

Sample Input 3

4 2 2 2
3
3 2
5 3
2 4

Sample Output 3

211.415926535897938

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TupleSections #-}
{-# LANGUAGE RankNTypes #-}
import qualified Data.ByteString.Lazy.Char8 as B
import Data.Maybe
import Data.List
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import Control.Monad
import Control.Monad.ST
-- import Debug.Trace
-- trace _ = id

solve :: Int -> Int -> Int -> Int -> Int -> [(Int,Int)] -> Double
solve xA yA xB yB bN lstXY =
  (fromIntegral ((xL+yL)*100)) + diffLen
  where
  cvtFX x | xA < xB = x - xA
          | otherwise = xA - x
  cvtFY y | yA < yB = y - yA
          | otherwise = yA - y
  cvtF (x,y) = (cvtFX x, cvtFY y)
  (xL, yL) = cvtF (xB, yB)
  vXY = VU.fromList $ sort $ filter eff $ map cvtF $ lstXY
    where eff (x,y) = 0 <= x && x <= xL && 0 <= y && y <= yL
  pt i = vXY VU.! i
  bM = VU.length vXY
  isUR (x,y) (x',y') = x+1 == x' && y+1 == y'
  allUR = and [ isUR (pt (i-1)) (pt i) | i <- [1..bM-1] ]
  isSpecCase1 = snd (pt 0) == 0 && snd (pt (bM-1)) == yL && allUR
  isSpecCase2 = fst (pt 0) == 0 && fst (pt (bM-1)) == xL && allUR
  diffLen
    | bM == 0 = 0
    | isSpecCase1 || isSpecCase2
      = (5*pi-20) * (fromIntegral (bM-1)) + (10*pi-20)
    | otherwise = (5*pi-20) * (fromIntegral dpHt)

  dpHt = runST $ VUM.new (bM+1) >>= dpSub
  dpSub vec = VU.foldM act 0 vXY where
    act ht (_,y) = do
      (idx, newHt) <- sch ht y
      VUM.write vec idx y
      return newHt

    sch 0 _ = return (1, 1)
    sch ht y = do
      yLast <- VUM.read vec ht
      if y > yLast then return (ht+1, ht+1)
                   else (,ht) <$> binsch 0 ht
      where
      binsch lo hi
        | lo + 1 == hi = return hi
        | otherwise = do
            let mid = (lo + hi) `div` 2
            yMid <- VUM.read vec mid
            if y < yMid then binsch lo mid else binsch mid hi

-------------------------------------------------------------------------------

readBInt :: B.ByteString -> Int
readBInt = fst . fromJust . B.readInt

tmain :: B.ByteString -> Double
tmain cont =
  let remLines0 = map B.words (B.lines cont)
      [bs_xA,bs_yA,bs_xB,bs_yB]:remLines1 = remLines0
      xA = readBInt bs_xA
      yA = readBInt bs_yA
      xB = readBInt bs_xB
      yB = readBInt bs_yB
      [bs_bN]:remLines2 = remLines1
      bN = readBInt bs_bN
      lstXY = map (\[x1,x2] -> (readBInt x1,readBInt x2)) remLines2
  in solve xA yA xB yB bN lstXY

outAnswer :: Double -> IO ()
outAnswer x = putStrLn $ show x

main :: IO ()
main = outAnswer . tmain =<< B.getContents


-------------------------------------------------------------------------------

inp1 = "1 1 6 5\n3\n3 2\n5 3\n2 4\n"
inp2 = "3 5 6 4\n3\n3 2\n5 3\n2 4\n"
inp3 = "4 2 2 2\n3\n3 2\n5 3\n2 4\n"
tv1 = tmain $ B.pack inp1
tv2 = tmain $ B.pack inp2
tv3 = tmain $ B.pack inp3
test1 = tv1 == 891.415926535897938
test2 = tv2 == 400.000000000000000
test3 = tv3 == 211.415926535897938
alltest = test1 && test2 && test3