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>
using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
#define REP(i,m,n) for(int i=(int)(m);i<(int)(n);++i)
#define rep(i,n) REP(i,0,n)
#define rREP(i,m,n) for(int (i)=(n)-1; (i)>=(m);--(i))
#define all(x) (x).begin(),(x).end()
#define out(y,x,h,w) (y)<0||(x)<0||(y)>=(h)||(x)>=(w)
constexpr int INF = 1000000001;
constexpr ll mod = 1000000007;
constexpr double PI = 3.1415926535897932;

template<class T>inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T>inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
inline void init() {
	cin.tie(nullptr); cout.tie(nullptr);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(15);
}

int main() {
	init();

	string s; cin >> s;
	bool is = false;
	rep(i, s.size() - 1) {
		if (s[i] == 'h') {
			if (s[i + 1] == 'i')is = true;
		}
	}
	cout << (is ? "Yes" : "No") << "\n";
		
	return 0;
}