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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define pa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
#define pb push_back

using ll = long long;
using namespace std;
using pint = pair<int,int>;
using pll = pair<ll,ll>;
const int INFint = 1e9;
const ll INFll = 1e18;
ll MOD=1e9+7;

ll gcd(ll a, ll b) { return __gcd(a,b); } //最大公約数
ll lcm(ll a, ll b) { return a/gcd(a,b)*b; } //最大公倍数

// aのn乗をMODで割りながら計算する
ll modpow(ll a, ll n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % MOD;
    a = a * a % MOD;
    n >>= 1; // right shift
  }
  return res;
}

// MODを法としたaの逆元を計算する
ll modinv(ll a) {
  return modpow(a, MOD - 2);
}

int main(){
  string s;
  cin>>s;
  for (int i(0);i<s.size()-1;i++){
    if (s[i] == 'h' && s[i+1] == 'i'){
      cout<< "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
  return 0;
}