Problem:
Score : 100 points

Problem Statement

You are given a string S as input. This represents a valid date in the year 2019 in the yyyy/mm/dd format. (For example, April 30, 2019 is represented as 2019/04/30.)

Write a program that prints Heisei if the date represented by S is not later than April 30, 2019, and prints TBD otherwise.

Constraints

S is a string that represents a valid date in the year 2019 in the yyyy/mm/dd format.

Input

Input is given from Standard Input in the following format:

S

Output

Print Heisei if the date represented by S is not later than April 30, 2019, and print TBD otherwise.

Sample Input 1

2019/04/30

Sample Output 1

Heisei

Sample Input 2

2019/11/01

Sample Output 2

TBD

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

#include <algorithm>
#include <cmath>
#include <complex>
#include <cstdio>
#include <iostream>
#include <iomanip>
#include <locale>
#include <map>
#include <numeric>
#include <queue>
#include <string>
#include <vector>

const long long int INF = 1e9;
typedef long long int ll;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define Rep(i, n) for(int i = 1; i <= (int)(n); i++)
#define llrep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define llRep(i, n) for(ll i = 1; i <= (ll)(n); i++)

using namespace std;

struct Point{
	int x,y;
};

bool vector_finder(std::vector<int> vec, int number) {
    auto itr = std::find(vec.begin(), vec.end(), number);
    size_t index = std::distance( vec.begin(), itr );
  if (index != vec.size()) { // 発見できたとき
    return true;
    }
  else { // 発見できなかったとき
    return false;
    }
}

ll factorial(ll N){
    ll ans = 1;
    llRep(i,N){
        ans *= i;
    }
    return ans;
}

ll extended_factorial(ll N){
    ll ans = 1;
    llRep(i,N){
        ans *= i;
        ans %= (ll)(1e9+7);
    }
    return ans;
}

vector<ll> Eratosthenes(ll N){
    bool arr[N+1];
    //initialization
    arr[0] = false;
    arr[1] = false;
	for(ll i = 2; i < N+1; i++){
		arr[i] = true;
	}
    //eratosthenes
	for(ll i = 2; i <= sqrt(N); i++){
		if(arr[i]){
			for(ll j = 0; i * (j + 2) <= N; j++){
				arr[i *(j + 2)] = false;//not prime number
			}
		}
	}
    vector<ll> prime;
    llRep(i,N){
        if (arr[i]==true){
            prime.push_back(i);
        }
    }
    return prime;
}

char alphabet[] = {
    'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'
};

// 二分探索
// llrep(i,N){
//     b = B[i];
//     auto cntA = lower_bound(A.begin(), A.end(), b);//イテレータなのでそのままllに変換とかはできない
//     auto cntC = upper_bound(C.begin(), C.end(), b);
//     ll cntA_ll = distance(A.begin(), cntA);
//     ll cntC_ll = C.size() - distance(C.begin(), cntC);
//     cnt += (cntA_ll * cntC_ll);
// }


int main(){
    string S;
    cin >> S;
    int date = stoi(S);
    if (date <= 20190430) cout << "Heisei" << endl;
    else cout << "TBD" << endl;

    return 0;
}