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: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <sstream>
#include <queue>
#include <deque>
#include <bitset>
#include <iterator>
#include <list>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <numeric>
#include <utility>
#include <climits>
#include <iomanip>
#include <unordered_map>
#include <unordered_set>
#include <time.h>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
using namespace std;
#define ll long long 
#define pb push_back
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
//typedef long long  MOD=1000000009;
ll MOD=998244353;
ll add(ll x, ll y) {ll res = x + y; return (res >= MOD ? res - MOD : res);}
ll mul(ll x, ll y) {ll res = x * y; return (res >= MOD ? res % MOD : res);}
ll sub(ll x, ll y) {ll res = x - y; return (res < 0 ? res + MOD : res);}
ll lcm(ll x, ll y) {ll res = (x * y) / __gcd(x, y); return res;}
ll power(ll x, ll y) {if (y < 0) return 1; ll res = 1; x %= MOD; while (y) {if (y & 1)res = mul(res, x); y >>= 1; x = mul(x, x);} return res;}

void iofoj(){
	#ifndef ONLINE_JUDGE
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	freopen("error.txt","w",stderr);
	#endif
}



void solution()
{
    ll k,i,j,n,m,w;
    string s;
    cin>>s;
    if(s[s.length()-1]=='s')
    {
        cout<<s<<"es";
    }
    else cout<<s<<"'s";
}

int main(){
    fio
    //iofoj();
    // ll t;
    // cin>>t;
    // while(t--)
    solution();
    return 0; 
}