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 Ruby (2.3.3) (judge verdict: Wrong Answer)

require 'set'

def main
    s = gets.chomp.split("")
    s.each_with_index do |c,i|
        if (i % 2 == 0 && c != 'h') || (i % 2 == 1 && c != 'i')
            puts "No"
            return
        end
    end
    puts "Yes"
end

main

## イディオムなど
# until s.empty? do
#
## 配列のインデックス
# s[0,6] # 0番目から6文字
# s[0..6] # 0番目から6番目まで
# s[5..-1] # 5番目から最後まで
#
## 全探索
# [0, 1].repeated_permutation(n) do |combination| # bit全探索
# array.combination(i) do |selected| # array から要素が i 個である部分列を全部抽出
#
## 無向グラフの幅優先探索(BFS)
# to = Array.new(n).map{Array.new()} # ある頂点と繋がる頂点のリスト
# m.times do
#     a,b = gets.chomp.split(" ").map(&:to_i)
#     to[a].push b
#     to[b].push a
# end
# q = Queue.new # BFS は queue
# ds = Array.new(n, -1)
# pre = Array.new(n, -1)
#
## その他
# a1 = Marshal.load(Marshal.dump(a2)) # 深いコピー
# matrix = Array.new(n).map{Array.new()} # 二次元配列の初期化
#