Problem:
Score : 100 points

Problem Statement

Two deer, AtCoDeer and TopCoDeer, are playing a game called Honest or Dishonest.
In this game, an honest player always tells the truth, and an dishonest player always tell lies.
You are given two characters a and b as the input. Each of them is either H or D, and carries the following information:

If a=H, AtCoDeer is honest; if a=D, AtCoDeer is dishonest.
If b=H, AtCoDeer is saying that TopCoDeer is honest; if b=D, AtCoDeer is saying that TopCoDeer is dishonest.

Given this information, determine whether TopCoDeer is honest.

Constraints

a=H or a=D.

b=H or b=D.

Input

The input is given from Standard Input in the following format:

a b

Output

If TopCoDeer is honest, print H. If he is dishonest, print D.

Sample Input 1

H H

Sample Output 1

H

In this input, AtCoDeer is honest. Hence, as he says, TopCoDeer is honest.

Sample Input 2

D H

Sample Output 2

D

In this input, AtCoDeer is dishonest. Hence, contrary to what he says, TopCoDeer is dishonest.

Sample Input 3

D D

Sample Output 3

H

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

#include <bits/stdc++.h>
using namespace std;
#define ll          long long
#define mp	    make_pair
#define pb          push_back
#define pii         pair<int,int>
#define vi          vector<int>
#define vii         vector<pii>
#define mi          map<int,int>
#define mii         map<pii,int>
#define all(a)      (a).begin(),(a).end()
#define x           first
#define y           second
#define sz(x)       (int)x.size()
#define endl        '\n'
#define MOD        1000000007
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define gcd(a,b)    __gcd((a),(b))
#define lcm(a,b)    ((a)*(b)) / gcd((a),(b))
#define rep(i,a,b)  for(int i=a;i<b;i++)
ll mult(ll a,ll b, ll p=MOD){return ((a%p)*(b%p))%p;}
ll multbig(ll a,ll b,ll mod){
    if(a==0 or b==0) return 0;
    if(a==1 or b==1) return (a*b)%mod;
    ll cur=multbig(a,b/2,mod);
    cur=(2*cur)%mod;
    if(b%2) cur=(cur+a)%mod;
    return cur;
}
ll add(ll a, ll b, ll p=MOD){return (a%p + b%p)%p;}
ll fpow(ll n, ll k, ll p = MOD) {ll r = 1; for (; k; k >>= 1LL) {if (k & 1LL) r = mult(r,n,p); n = mult(n,n,p);} return r;}
ll inv(ll a, ll p = MOD) {return fpow(a, p - 2, p);}
ll inv_euclid(ll a, ll m = MOD){ll m0 = m;ll y = 0, x = 1;if (m == 1)return 0;while (a > 1){ll q = a / m;ll t = m;m = a % m, a = t;t = y;y = x - q * y;x = t;}if (x < 0)x += m0;return x;}
#define N	55
void solve(){
	char a, b;
	cin>>a>>b;
	if(a=='H')
		cout<<b<<endl;
	else{
		if(b=='H') cout<<a<<endl;
		else cout<<'H'<<endl;
	}
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
     t =1;
//     cin>>t;
    while(t--){
        solve();
    }
    return 0;
}