#include<iostream>
#include <string>
#include <vector>
using namespace std;
class DoubleLetter {public:string ableToSolve(string S) {while (1){int flag=0;for (int i=1;i<S.size();i++)if (S[i]==S[i-1]){S.erase(i-1,2);flag=1;}if (!flag||S.size()==0) break;}if (S.size()==0) return "Possible";else return "Impossible";}
};// Powered by FileEdit
// Powered by TZTester 1.01 [25-Feb-2003]
// Powered by CodeProcessor

#include<iostream>
#include <string>
#include <vector>
#include<string.h>
using namespace std;
int b[12345];
int dis[12][12];
int mp[12][12];class Egalitarianism3Easy {public:int maxCities(int n, vector <int> a, vector <int> b, vector <int> len) {memset(dis,0x3f3f3f,sizeof(dis));for (int i=1;i<=n;i++) dis[i][i]=0;for (int i=0;i<a.size();i++) dis[a[i]][b[i]]=dis[b[i]][a[i]]=len[i];for (int k=1;k<=n;k++)for (int i=1;i<=n;i++)for (int j=1;j<=n;j++)dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);int Max=1;for (int i=1;i<(1<<n);i++){vector<int> V;V.clear();for (int j=0;j<n;j++)if (((i>>j)&1)==1) V.push_back(j+1);int m=V.size();int same=dis[V[0]][V[m-1]];int flag=1;for (int k=0;k<m;k++)for (int j=0;j<m;j++)if (k!=j&&dis[V[k]][V[j]]!=same){flag=0;break;}if (flag) Max=max(Max,m);}return Max;}
};

#include<iostream>
#include <string>
#include <vector>
#include<string.h>
#include<algorithm>
using namespace std;
class SuffixArrayDiv2 {public:string  arr[100];int p[100],a[100];//注意P:表示字符串排完序后，所对应原来数组的下标//    A:表示原先字符串在排序后的下标// 这里很绕，熟悉后缀数组的应该了解char c[100];string smallerOne(string s) {for (int i=0;i<s.size();i++)arr[i]=&s[i];//保存后缀sort(arr,arr+s.size());int len=s.size();for (int i=0;i<s.size();i++){p[i]=len-arr[i].size();a[len-arr[i].size()]=i;}a[len]=-1;c[p[0]]='a';for (int i=1;i<len;i++){if (a[p[i-1]+1]<a[p[i]+1]) c[p[i]]=c[p[i-1]];//这个表示：我们确定位置为p[i]的字符，//如果当前字符串s1,与后面的字符串s2（都是排序后的字符串）//比较它们后面+1的字符串的大小。这里可以忽视我的解释else c[p[i]]=c[p[i-1]]+1;}string ans;for (int i=0;i<len;i++) ans+=c[i];if (ans<s) return "Exists";else return "Does not exist";}
};int main()
{SuffixArrayDiv2 b;string s;cin>>s;cout<<b.smallerOne(s);return 0;
}