#include <bits/stdc++.h>
#include <iomanip>
#include <math.h>
//#include <unistd.h>
//#include <stdio.h>
#define rep(i,x) for(ll i = 0;i<x;i++)
#define all(a) (a).begin(),(a).end()
using ll = long long;
using ld = long double;
using namespace std;
using dou = double;
const ll inf = 2147483647;
const ll INF = 1LL << 60;
const ld pi = 3.14159265358;
const ll mod = 1000000007;
typedef pair<ll,ll> P;
using graph = vector<vector<ll>>;
template<class T> inline bool chmax(T& a, T b){if(a < b){a = b; return true;}return false;}
template<class T> inline bool chmin(T& a, T b){if(a > b){a = b; return true;}return false;}
ll gcd(ll a,ll b){
    if(a < b)swap(a,b);
    if(a % b == 0)return b;
    else return gcd(b,a%b);
}
//素数判定O(sqrt(N))
bool isp(ll n){
    bool res = true;
    if(n == 1)return false;
    else{
        for(ll i = 2;i * i <= n;i++){
            if(n % i == 0){
                res = false;
                break;
            }
        }
        return res;
    }
}

ll fac[510000], finv[510000], inv[510000];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < 510000; i++){
        fac[i] = fac[i - 1] * i % mod;
        inv[i] = mod - inv[mod%i] * (mod / i) % mod;
        finv[i] = finv[i - 1] * inv[i] % mod;
    }
}

// 二項係数計算
ll nCk(ll n, ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
}
//nのm乗をMODで割ったあまりO(logN)
ll modpow(ll n,ll m,ll MOD){
    ll res = 1;
    while(m){
        if(m & 1)res = (res * n) % MOD;
        m >>= 1;
        n *= n;
        n %= MOD;
    }
    return res;
}

//printf("%.10f\n", N);
/*  vector<ll> vec = {1,2,5,6,8,9,10};
    cout << binary_search(all(vec),5) << endl; -> true*/
/*
    vector<ll> vec = {1,2,5,6,8,9,10};
    auto n = upper_bound(all(vec),4);
    cout << *n << endl; -> 5*/

//    cout << fixed << setprecision(15);

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll S;
    cin >> S;
    if(S % 111 == 0)
    cout << S / 111 * 111 << endl;
    else cout << (S / 111 + 1) * 111 << endl;
}
//  Created by conan1024hao in 2019.
//  Copyright © 2019 conan1024hao. All rights reserved.
//  専用ライブラリです、自由にコピーして構いません。
//  感谢看我的代码！Wechat:conan1024hao Twitter/QQ:810396815
#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 10e8
#define INF 10e16
#define MOD 1000000007
#define mod 1000000007
#define rep(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
//using Int=int_fast64_t;
typedef long long int ll;
typedef pair<ll,ll>Pll;
typedef pair<int,int>Pin;
long long GCD(long long a, long long b) { return b ? GCD(b, a%b) : a; }
long long LCM(long long a, long long b)  {return a/GCD(a,b)*b;}
int dx[8]={-1,0,1,0,1,1,-1,-1};
int dy[8]={0,-1,0,1,1,-1,1,-1};
char dir[4]={'u','l','d','r'};
ll cmp1(pair<ll,string>a,pair<ll,string> b){
        if(a.fi!=b.fi)
        return a.fi<b.fi;
        else
        return a.se<b.se;
}

//---------------------------------------------------------------------------

//---------------------------------------------------------------------------
int main(){//問題をちゃんと見ろ!!!!!!!!!!!!!!!!!      llか??????????        memset()!!!!!!!!!!!!       ペナを減らせ!!!!!!!!!!!!!
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------
    string s;cin>>s;
    if(s.length()==2)
        cout<<s<<Endl;
    else{
        reverse(all(s));
        cout<<s<<Endl;
    }
    //-------------------------------
    return 0;
}
//---------------------------------------------------------------------------







