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 <algorithm>
#include <bits/stdc++.h>
#include <bitset>
#include <cassert>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <math.h>
#include <memory>
#include <queue>
#include <set>
#include <stack>
#include <utility>
#include <vector>
using namespace std;
#define REP(i, n) for (int64_t i = 0; i < (int64_t)n; i++)
#define REP_FOR(i, a, b) for (int64_t i = (int64_t)a; i < (int64_t)b; i++)
#define REP_REV(i, a, b) for (int64_t i = (int64_t)a; i > (int64_t)b; i--)
#define FORALL(v) (v).begin(), (v).end()
#define mp(a, b) make_pair(a, b)
#define eb(x) emplace_back(x)
#define F first
#define S second
typedef long long ll;
typedef long double ld;
typedef pair<int64_t, int64_t> pii;
typedef pair<ll, ll> pll;
typedef pair<ld, ld> pdd;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int64_t> vii;
typedef vector<vii> vvii;
const ll mod = 1e9 + 7;
const int64_t infi = 1147483600;
const ll infl = 4e18 + 5;
const char EOL = '\n';
//cout << fixed << setprecision(17) << res << endl;
const ll MOD = 998244353;
 
int main() {
    string s;
    cin >> s;
    auto n = s.size();
    if (n % 2 == 1) {
        cout << "No" << endl;
        return 0;
    }
    for (ll i = 0; i < n; i = i + 2) {
        if (s[i] != 'h' || s[i+1] != 'i') {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;
}