When preparing a tournament, Codeforces coordinators try treir best to make the first problem as easy as possible. This time the coordinator had chosen some problem and asked nn people about their opinions. Each person answered whether this problem is easy or hard.
If at least one of these nn people has answered that the problem is hard, the coordinator decides to change the problem. For the given responses, check if the problem is easy enough.
Input The first line contains a single integer nn (1≤n≤1001≤n≤100) — the number of people who were asked to give their opinions.
The second line contains nn integers, each integer is either 00 or 11. If ii-th integer is 00, then ii-th person thinks that the problem is easy; if it is 11, then ii-th person thinks that the problem is hard.
Output Print one word: “EASY” if the problem is easy according to all responses, or “HARD” if there is at least one person who thinks the problem is hard.
You may print every letter in any register: “EASY”, “easy”, “EaSY” and “eAsY” all will be processed correctly.
Examples Input 3 0 0 1 Output HARD Input 1 0 Output EASY Note In the first example the third person says it’s a hard problem, so it should be replaced.
In the second example the problem easy for the only person, so it doesn’t have to be replaced. 分析：一开始wa的原因是没有在hard后跳出
#include<iostream> #include<cmath> #include<algorithm> using namespace std; typedef long long ll; int main() { int n,p,k=0; cin>>n; for(int i=1;i<=n;i++) { cin>>p; if(p==0) k++; else {cout<<"HARD";break;} } if(k==n) cout<<"EASY"; }