#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
typedef long long ll;
const int INF = 1<<30;
const ll INFLL = 1LL<<60;
const ll MOD = 998244353;
const double INFD = 1.0E10;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, -1, 0, 1};
map<int, char> dm = {{0, 'D'}, {1, 'L'}, {2, 'U'}, {3, 'R'}};
// const int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
// const int dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
using Pair = pair<ll, ll>;
using mint = atcoder::modint998244353;
// using mint = atcoder::modint1000000007;


string solve(string s){
    int n = s.size();
    
    if (s.front() <= '3'){
        string ans(n - 1, '5');
        return ans;
    }
    else if (s.front() == '4'){
        bool flag = false;
        for (int i = 0; i < n; i++){
            if (!flag && s[i] < '4'){
                string ans(n - 1, '5');
                return ans;
            }
            if (s[i] >= '5') flag = true;
        }
    }
    
    //これ以降はその桁のを構築できる。
    int flag = -1;
    for (int i = 0; i < n; i++){
        if (flag == INF){
            s[i] = '5';
            continue;
        }
        
        if (s[i] >= '5') flag = i;
        else if (s[i] < '4'){
            s[flag] = (s[flag] >= '6' ? '5' : '4');
            for (int j = flag + 1; j <= i; j++) s[j] = '5';
            flag = INF;
        }
    }
        
    return s;
}




int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
    
    string s; cin >> s;
    string ans = solve(s);
    cout << ans << endl;
    
    return 0;
}