Problem:
Score : 100 points

Problem Statement

In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.

In Taknese, the plural form of a noun is spelled based on the following rules:

If a noun's singular form does not end with s, append s to the end of the singular form.

If a noun's singular form ends with s, append es to the end of the singular form.

You are given the singular form S of a Taknese noun. Output its plural form.

Constraints

S is a string of length 1 between 1000, inclusive.

S contains only lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

Print the plural form of the given Taknese word.

Sample Input 1

apple

Sample Output 1

apples

apple ends with e, so its plural form is apples.

Sample Input 2

bus

Sample Output 2

buses

bus ends with s, so its plural form is buses.

Sample Input 3

box

Sample Output 3

boxs

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
  
using namespace std;
 
#define endl "\n"
#define JAM(x) cout<< #x <<" = "<<x<<"\n"

 
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef pair<int,int> PII;
typedef vector<PII> VPI;
 
const int INF=2e9;
const int MX=100005;
const int MOD=1000000007;
const double PI=acos(-1.0);



long long int power(long long int a,long long int b)
{
    a %= MOD;
    long long int res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return res;
}
bool isprime(int n)
{
    if(n==1)
    return 0;
    int i;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        return false;
    }
    return true;
}




/*_______________________________________________*/





int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    string s;
    cin>>s;
    int n=s.length();
    if(s[n-1]=='s')
    s+="es";
    else
    s+="s";
    cout<<s;
    
}