Problem:
Score : 100 points

Problem Statement

A Hitachi string is a concatenation of one or more copies of the string hi.

For example, hi and hihi are Hitachi strings, while ha and hii are not.

Given a string S, determine whether S is a Hitachi string.

Constraints

The length of S is between 1 and 10 (inclusive).

S is a string consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S is a Hitachi string, print Yes; otherwise, print No.

Sample Input 1

hihi

Sample Output 1

Yes

hihi is the concatenation of two copies of hi, so it is a Hitachi string.

Sample Input 2

hi

Sample Output 2

Yes

Sample Input 3

ha

Sample Output 3

No

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

import           Control.Monad
import           Control.Monad.ST
import qualified Data.Array                    as A
import qualified Data.Array.Unboxed            as AU
import qualified Data.Array.ST                 as ST
import qualified Data.ByteString.Char8         as BS
import           Data.Char
import           Data.Foldable                  ( toList )
import           Data.Int
import qualified Data.IntMap                   as IM
import           Data.List
import qualified Data.Map                      as M
import           Data.Maybe
import           Data.Ord
import qualified Data.Set                      as S
import qualified Data.Sequence                 as SQ
import           Data.Time
import qualified Data.Vector                   as V
import qualified Data.Vector.Mutable           as VM
import qualified Data.Vector.Unboxed           as VU
import qualified Data.Vector.Unboxed.Mutable   as VUM

readInt = map (fst . fromJust . BS.readInt) . BS.words <$> BS.getLine
readInts n = replicateM n readInt
readString = map BS.unpack . BS.words <$> BS.getLine
readStrings n = replicateM n readString
toPair [x, y] = (x, y)

main = do
  [s] <- readString
  putStrLn $ solve s

solve s | f s       = "Yes"
        | otherwise = "No"

f []  = True
f [_] = False
f (s : s' : ss) | [s, s'] == "hi" = f ss
                | otherwise       = False