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 <iostream>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <tuple>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cassert>
#include <cstdint>
#include <cctype>
#include <numeric>
#include <bitset>
#include <functional>

using namespace std;

using ll =  long long;
using Pll = pair<ll, ll>;
using Pii = pair<int, int>;

constexpr int INF = 1 << 30;
constexpr ll LINF = 1LL << 60;
constexpr ll MOD = 1000000007;
constexpr long double EPS = 1e-10;
constexpr int dyx[4][2] = {
    { 0, 1}, {-1, 0}, {0,-1}, {1, 0}
};

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    string s;
    cin  >> s;
    int n = s.length();
    if(n%2) {
        cout << "No" << endl;
        return 0; 
    }
    for(int i=0;i<n;++i) {
        if(i%2 == 0) {
            if(s[i] != 'h') {
                cout << "No" << endl;
                return 0;
            }
        } else {
            if(s[i] != 'i') {
                cout << "No" << endl;
                return 0;
            }
        }
    }

    cout << "Yes" << endl;
    return 0;
}