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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define int long long int
#define mod 1000000007
#define string_mod 2549536629329_base_255
#define pb push_back
#define F first
#define S second
#define endl "\n"
#define ss second
#define all(v) v.begin(), v.end()
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
int power(int x, unsigned int y, int p)
{
	int res = 1;
	x = x % p;
	while (y > 0)
	{
		if (y & 1)
			res = (res * x) % p;
		y = y >> 1;
		x = (x * x) % p;
	}
	return res % p;
}
int lcm(int a, int b)
{
	return a / __gcd(a, b) * b;
}
void solve()
{
	string s;
	cin >> s;
	if(s[s.size()-1]=='s'){
		s += "es";
	}
	else{
		s += "s";
	}
	cout << s << endl;
}
int32_t main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int tc;
	tc = 1;
	//cin >> tc;
	while (tc--)
	{
		solve();
	}

	return 0;
}