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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, n) for (int i = (int)(n); i >= 0; i--)
#define REP(i, m, n) for (int i = (int)(m); i <= (int)(n); i++)
#define all(v) v.begin(), v.end()
typedef long long ll;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const long long INF = 1LL << 60;

int main(){
  string s;
  cin >> s;

  bool ok = true;
  if(s.size()%2 != 0){
    cout << "No" << endl;
    return 0;
  }
  rep(i, s.size()){
    if(i%2 == 0){
      if(s[i] != 'h'){
        ok = false;
      }
    }
    else{
      if(s[i] != 'i'){
        ok = false;
      }
    }
  }

  cout << (ok ? "Yes" : "No") << endl;

  return 0;
}