#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
//回文日期要以年为划分，枚举所有的回文日期再判断的方式更加简便
const int N = 9999;//要遍历所有比输入年份大的情况，那么就是从输入年份到9999年，不是到9000年！
int months[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
int nyear,nmonth,nday;
bool check1(int year,int month,int day){
    //cout<<year<<" "<<month<<" "<<day<<endl;
    if(month<1||month>12||day>31){
        return false;
    }else{
        //cout<<year<<" "<<month<<" "<<day<<endl;
        int flag = 0;
        if(month==2){
            if((year%4==0&&year%100)||(year%400==0)){
                flag = 1;
            }
        }
        if(day>months[month]+flag){
            //cout<<year<<" "<<month<<" "<<day<<endl;
            return false;
        }

        if((year==nyear&&month<nmonth)||(year==nyear&&month==nmonth&&day<=nday)){
            return false;
        }

    }
    return true;
}
bool check2(int year,int month,int day){
    bool flag = check1(year,month,day);
    if(flag){
        //cout<<year<<" "<<month<<" "<<day<<endl;
        return true;
    }
}
int seek(int n){
    nyear = n/10000;
    nmonth = (n%10000)/100;
    nday = n%100;
    for(int i = nyear;i<=N;i++){
        int year = i;
        int month = (year%10)*10+(year/10)%10;
        int day = ((year/100)%10)*10+year/1000;
        bool ans1 = check1(year,month,day);
        if(ans1){
            int num1 = year*10000 + month*100+day;
            cout<<num1<<endl;
            break;
        }
    }
    for(int i = nyear;i<=N;i++){
        int year = i;
        int month = (year%10)*10+(year/10)%10;
        int day = ((year/100)%10)*10+year/1000;
        if(year/100==year%100&&(year%10!=(year/10)%10)){//如果是ABAB一定是BABA的，并且要保证A≠B
            bool ans2 = check2(year,month,day);
            if(ans2){
                int num2 = year*10000 + month*100+day;
                cout<<num2<<endl;
                break;
            }
        }

    }
}
int main(){
    int n ;
    cin>>n;
    seek(n);
    return 0;
}