Technocup 2019 - Elimination Round 1.A. In Search of an Easy Problem

本文介绍了一个简单的算法,用于根据多名参与者的意见评估比赛问题的难度。如果至少有一名参与者认为问题困难,则该问题将被标记为硬问题。文章通过示例说明了如何使用此算法进行问题难度的判断。

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A. In Search of an Easy Problem

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

3
0 0 1

output

Copy

HARD

input

Copy

1
0

output

Copy

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.

#include<bits/stdc++.h>
using namespace std;
#define maxn 300005
#define ll long long
#define M(a) memset(a,0,sizeof(a))

int main()
{
    int t,n;
    int b[10];

    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    while(cin>>n)
    {
        int f=0,a;
        for(int i=0;i<n;i++){
            cin>>a;
            if(a==1) f=1;
        }
        if(f) cout<<"HARD"<<endl;
        else cout<<"EASY"<<endl;
    }
}

 

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