Problem:
Score : 600 points

Problem Statement

A subsequence of a string S is a string that can be obtained by deleting zero or more characters from S without changing the order of the remaining characters.
For example, arc, artistic and (an empty string) are all subsequences of artistic; abc and ci are not.

You are given a string A consisting of lowercase English letters.
Find the shortest string among the strings consisting of lowercase English letters that are not subsequences of A.
If there are more than one such string, find the lexicographically smallest one among them.

Constraints

1 \leq |A| \leq 2 \times 10^5

A consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

A

Output

Print the lexicographically smallest string among the shortest strings consisting of lowercase English letters that are not subsequences of A.

Sample Input 1

atcoderregularcontest

Sample Output 1

b

The string atcoderregularcontest contains a as a subsequence, but not b.

Sample Input 2

abcdefghijklmnopqrstuvwxyz

Sample Output 2

aa

Sample Input 3

frqnvhydscshfcgdemurlfrutcpzhopfotpifgepnqjxupnskapziurswqazdwnwbgdhyktfyhqqxpoidfhjdakoxraiedxskywuepzfniuyskxiyjpjlxuqnfgmnjcvtlpnclfkpervxmdbvrbrdn

Sample Output 3

aca

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define rep(i,a,n) for(int i=(a);i<=(n);++i)
#define dep(i,a,n) for(int i=(a);i>=(n);--i)
#define eps 1e-8
#define pi 3.1415926535897
#define sqr(x) ((x)*(x))
#define MAX(a,b) a=max(a,b)
#define MIN(a,b) a=min(a,b)
#define SZ(x) ((int)(x).size())
#define CPY(a,b) memcpy(a,b,sizeof(a))
#define CLR(a) memset(a,0,sizeof(a))
#define POSIN(x,y) (1<=(x)&&(x)<=n&&1<=(y)&&(y)<=m)
#define all(x) (x).begin(),(x).end()
#define COUT(S,x) cout<<fixed<<setprecision(x)<<S<<endl
#define buli(x) (__builtin_popcountll(x))
#define bur0(x) (__builtin_ctzll(x))
#define bul2(x) (63-__builtin_clzll(x))
#define pw(x) ((ll(1))<<(x))
#define upmo(a,b) (((a)=((a)+(b))%mo)<0?(a)+=mo:(a))
#define mmo(a,b) (((a)=1ll*(a)*(b)%mo)<0?(a)+=mo:(a))
#define y0 y0z
#define y1 y1z
#define yn ynz
#define j0 j0z
#define j1 j1z
#define jn jnz
#define tm tmz
#ifdef LOCAL
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...) 
#endif
typedef long long ll;
typedef double lf;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<lf,lf> pff;
typedef complex<double> CD;
const int inf=0x3f3f3f3f;
const int mo=1000000007;
inline void gn(long long&x){
	int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
	while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
inline void gn(double&x){double t;scanf("%lf",&t);x=t;}
inline void gn(long double&x){double t;scanf("%lf",&t);x=t;}
template<class T1,class T2>inline void gn(T1&r,T2&s){gn(r),gn(s);}
template<class T1,class T2,class T3>inline void gn(T1&r,T2&s,T3&t){gn(r),gn(s),gn(t);}
template<class T1,class T2,class T3,class T4>inline void gn(T1&r,T2&s,T3&t,T4&u){gn(r),gn(s),gn(t),gn(u);}
inline void gs(char *s){scanf("%s",s);}
inline void gc(char &c){while((c=getchar())>126 || c<33);}
inline void pc(char c){putchar(c);}
const int DX[]={1,0,-1,0},DY[]={0,1,0,-1};
ll powmod(ll a,ll b) {ll res=1;a%=mo;for(;b;b>>=1){if(b&1)res=res*a%mo;a=a*a%mo;}return res;}
ll powmod(ll a,ll b,ll mo) {ll res=1;a%=mo;for(;b;b>>=1){if(b&1)res=res*a%mo;a=a*a%mo;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//*******************************************

const int N=211111,M=111111;
int l,m,n,t,C,tot;
char s[N];
int a[N],ok[N];
int main(){
#ifdef LOCAL
	freopen("E.in","r",stdin);//freopen("E.out","w",stdout);
#endif
	scanf("%s",s+1);n=strlen(s+1);
	int cnt=0;
	dep(i,n,1){
		if(a[s[i]-'a'])continue;
		a[s[i]-'a']=1;
		++cnt;
		if(cnt==26){
			rep(i,0,25)a[i]=0;
			cnt=0;
			ok[++tot]=i;
		}
	}
	ok[0]=n+1;
	int q=1;
	dep(p,tot,0){
		rep(i,0,25)a[i]=0;
		rep(i,q+1,ok[p]-1)a[s[i]-'a']=1;
		int o;
		rep(i,0,25)if(!a[i]){pc(i+'a');o=i;break;}
		if(p==0)break;
		++q;while(s[q]!=o+'a')++q;
	}
	puts("");
	return 0;
}