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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define double long double
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(),v.end()
#define pb push_back
#define eb emplace_back
template<class T>void chmax(T &a,T b){if(a<b)a=b;}
template<class T>void chmin(T &a,T b){if(a>b)a=b;}
constexpr int INF=1000000000000000000;
constexpr int mod=1000000007;
constexpr double eps=1e-8;
const double pi=acos(-1);
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int Random(int mi,int ma){
    random_device rnd;
    mt19937 mt(rnd());//32bit
    //[mi,ma]
    uniform_int_distribution<int>engine(mi,ma);
    return engine(mt);
}
int gcd(int a,int b){
    if(b==0)return a;
    return gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
bool prime(int a){
    if(a==1)return false;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)return false;
    }
    return true;
}
int modpow(int a,int b){
    if(b==0)return 1;
    if(b%2)return modpow(a,b-1)*a%mod;
    int memo=modpow(a,b/2);
    return memo*memo%mod;
}
vector<int>kaijo,invkaijo;
void init_fact(int n){
    kaijo.resize(n+1);
    invkaijo.resize(n+1);
    kaijo[0]=1;
    for(int i=1;i<=n;i++){
        kaijo[i]=kaijo[i-1]*i;
        kaijo[i]%=mod;
    }
    rep(i,n+1)invkaijo[i]=modpow(kaijo[i],mod-2);
}
int comb(int a,int b){
    if(a<b)return 0;
    if(a<0||b<0)return 0;
    return kaijo[a]*modpow(kaijo[a-b],mod-2)%mod*modpow(kaijo[b],mod-2)%mod;
}
int inv(int x){
    x=modpow(x,mod-2);
    return x;
}
signed main(){
    cin.tie(0);ios::sync_with_stdio(false);
    string s,ans="";
    cin>>s;
    int n=s.size(),dp[200010],pos=0,nxt[200010][26];
    vector<int>v[26];
    rep(i,n)v[s[i]-'a'].pb(i);
    rep(i,26)v[i].pb(n);
    fill(dp,dp+n,INF);
    dp[n]=1;
    for(int i=n-1;i>=0;i--)rep(j,26)chmin(dp[i],dp[*lower_bound(all(v[j]),i)+1]+1);
    return 0;
    while(pos<n){
        rep(i,26){
            int nxt=*lower_bound(all(v[i]),pos);
            if(dp[pos]==dp[nxt+1]+1){
                ans+='a'+i;
                pos=nxt+1;
                break;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}