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 ll long long
#define endl "\n"
#define speed ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define pb push_back
#define F first
#define S second
#define um map <ll,ll>
#define rep(i,z,n)  for(int i=z;i<n;i++)
#define repi(i,z,n) for(int i=z;i<=n;i++)
#define repn(i,z,n) for(int i=n-1;i>=z;i--)
#define vec vector<ll>
#define vecp vector<pair<ll,ll>>
#define pi (double)3.14159265358979323846
#define ld long double
#define all(z) z.begin(),z.end()

ll power(ll a, ll b, ll m) { ll ans = 1; a = a % m; if (a == 0) return 0; while (b) { if (b & 1) ans = (ans * a) % m; b /= 2; a = (a * a) % m; } return ans; }
ll modInverse(ll a, ll m) {return power(a, m - 2, m);}
ll binpow(ll a, ll b) {ll res = 1; while (b > 0) {if (b & 1)res = res * a; a = a * a; b >>= 1;} return res;}
ll binlog2(ll n) {ll logValue = -1; while (n) {logValue++; n >>= 1;} return logValue;}
ll ceil(ll t1, ll t2) {ll ans = (t1 + t2 - 1) / t2; return ans;}
void print(vec a) {rep(i, 0, a.size())cout << a[i] << " "; cout << endl;}
const ll mod = 1e9 + 7;
const ll N = 2e5 + 5;
const ll inf = 2e18;


void solve() {
    string s;
    cin >> s;
    if (s.back() == 's') {
        cout << s + "es" << endl;
    }
    else {
        cout << s + "s" << endl;
    }
}


int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    speed;
    int test = 1;
    // cin >> test;
    while (test--) {
        // cout << "Case #" << t1 << ": ";
        solve();
    }
    return 0;
}