// Slower C++ solution:
#include <vector>

#include <map>

#include <set>

#include <stack>

#include <queue>

#include <algorithm>

#include <numeric>

#include <sstream>

#include <iostream>

#include <iomanip>

#define _USE_MATH_DEFINES

#include <cmath>

#include <climits>

#include <cstdlib>

#include <cstring>

#include <cfloat>

#include <cmath>

#include <ctime>

#include <cstdio>

#include <cassert>

using namespace std;



#define EPS 1e-12

#define ull unsigned long long

#define ll long long

#define VI vector<ll>

#define PII pair<ll, ll> 

#define VVI vector<vector<ll> >

#define REP(i,n) for(int i=0,_n=(n);(i)<(int)_n;++i)

#define RANGE(i,a,b) for(int i=(int)a,_b=(int)(b);(i)<_b;++i)

#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)

#define ALL(c) (c).begin(), (c).end()

#define ALLR(c) (c).rbegin(), (c).rend()

#define PB push_back

#define MP(a, b) make_pair(a, b)

#define POPCOUNT __builtin_popcount

#define POPCOUNTLL __builtin_popcountll

#define CLEAR(table, v) memset(table, v, sizeof(table));

#define PRINT1(table, D0) REP(d0, D0) cout<<table[d0]<<" "; cout<<"\n";

#define PRINT2(table, D0, D1) REP(d0, D0) { REP(d1, D1) cout<<table[d0][d1]<<" "; cout<<"\n"; }

#define PRINT3(table, D0, D1, D2) REP(d0, D0) { REP(d1, D1) { REP(d2, D2) cout<<table[d0][d1][d2]<<" "; cout<<"\n"; } cout<<"\n"; }

#define UNIFORM_DOUBLE(a, b) (((b-a)*(double)rand()/RAND_MAX)+a) // [a, b) 

#define UNIFORM_LL(a, b) (ll)UNIFORM_DOUBLE(a, b) // [a, b) 

#define IN(v, lo, hi) ((lo)<=(v) && (v)<(hi))

#define DD(v) cout<<#v<<": "<<v<<endl

template <typename T0, typename T1> std::ostream& operator<<(std::ostream& os, const map<T0, T1>& v) { for( typename map<T0, T1>::const_iterator p = v.begin(); p!=v.end(); p++ ){os << p->first << ": " << p->second << " ";} return os; }

template <typename T0, typename T1> std::ostream& operator<<(std::ostream& os, const pair<T0, T1>& v) { os << v.first << ": " << v.second << " "; return os; }

template <typename T> std::ostream& operator<<(std::ostream& os, const vector<T>& v) { for( int i = 0; i < (int)v.size(); i++ ) { os << v[i] << " "; } return os; }

template <typename T> std::ostream& operator<<(std::ostream& os, const vector<vector<T> >& v) { for( int i = 0; i < (int)v.size(); i++ ) { os << v[i] << endl; } return os; }

template <typename T> std::ostream& operator<<(std::ostream& os, const set<T>& v) { vector<T> tmp(v.begin(), v.end()); os << tmp; return os; }

template <typename T> std::ostream& operator<<(std::ostream& os, const deque<T>& v) { vector<T> tmp(v.begin(), v.end()); os << tmp; return os; }



#define MOD 1000000007LL

#define INF (1LL<<60)



ll CS = 26;

//ll CS = 3;



ll N;

string s;

VVI nxt;



// i以降でcが最初に出てくる位置

ll pos(ll i, ll c) {

//	cout<<"pos "<<i<<" "<<c<<endl;

	return nxt[c][i];

//	RANGE(j, i, N) if(s[j]=='a'+c) return j;

//	return -1;

}



using plll = pair<int, pair<int, int>>;



vector<vector<plll>> memo;

//map<pair<ll, ll>, pair<ll, pair<ll, ll>>> memo;



// iにいてcを選んだ時の残りstep

plll f(int i, int c) {

//	cout<<"f "<<i<<" "<<c<<endl;

	// 範囲外なのでcでOK

	if(i>=N) return MP(1, MP(-1, -1));



	if(memo[c][i].first!=-1) return memo[c][i];



	ll p = pos(i, c);

//	DD(p);

	if(p==-1) return memo[c][i] = MP(1, MP(-1, -1));

	auto lans = MP(1<<30, MP(-1, -1));

	REP(j, CS) {

		auto a = f(p+1, j);

		if(1+a.first < lans.first) {

			lans = MP(1+a.first, MP(p+1, j));

		}

	}

//	cout<<"f "<<i<<" "<<c<<" -> "<<lans<<endl;

	return memo[c][i] = lans;

}



int main() {

	cin.tie(0);

	ios::sync_with_stdio(false);

	while(cin>>s) {

		N = s.size();

		memo = vector<vector<plll>>(CS, vector<plll>(N, MP(-1, MP(0, 0))));

		nxt = VVI(CS, VI(N+1, -1));

		for(int i=N-1;i>=0;i--) {

			REP(j, CS) if(s[i]-'a'==j) nxt[j][i] = i; else if(i+1<N) nxt[j][i] = nxt[j][i+1];

		}

		auto mv = MP(1<<30, MP(0, 0));

		ll first = -1;

		REP(j, CS) {

			auto vv = f(0, j);

			if(vv.first < mv.first) {

				mv = vv;

				first = j;

			}

		}

//		DD(mv);

//		DD(first);

		string ans;

		ll p = 0;

		ll c = first;

		while(1) {

			ans.PB('a'+c);

			auto v = f(p, c);

			p = v.second.first;

			c = v.second.second;

			if(p==-1) break;

		}

		REP(i, N) REP(j, CS) {

//			DD(i);DD(j);DD(f(i, j));

		}

//		DD(ans);

		cout<<ans<<endl;



//		break;

	}

	

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.456014758570644x):
#include <vector>

#include <map>

#include <set>

#include <stack>

#include <queue>

#include <algorithm>

#include <numeric>

#include <sstream>

#include <iostream>

#include <iomanip>

#define _USE_MATH_DEFINES

#include <cmath>

#include <climits>

#include <cstdlib>

#include <cstring>

#include <cfloat>

#include <cmath>

#include <ctime>

#include <cstdio>

#include <cassert>

using namespace std;



#define EPS 1e-12

#define ull unsigned long long

#define ll long long

#define VI vector<ll>

#define PII pair<ll, ll> 

#define VVI vector<vector<ll> >

#define REP(i,n) for(int i=0,_n=(n);(i)<(int)_n;++i)

#define RANGE(i,a,b) for(int i=(int)a,_b=(int)(b);(i)<_b;++i)

#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)

#define ALL(c) (c).begin(), (c).end()

#define ALLR(c) (c).rbegin(), (c).rend()

#define PB push_back

#define MP(a, b) make_pair(a, b)

#define POPCOUNT __builtin_popcount

#define POPCOUNTLL __builtin_popcountll

#define CLEAR(table, v) memset(table, v, sizeof(table));

#define PRINT1(table, D0) REP(d0, D0) cout<<table[d0]<<" "; cout<<"\n";

#define PRINT2(table, D0, D1) REP(d0, D0) { REP(d1, D1) cout<<table[d0][d1]<<" "; cout<<"\n"; }

#define PRINT3(table, D0, D1, D2) REP(d0, D0) { REP(d1, D1) { REP(d2, D2) cout<<table[d0][d1][d2]<<" "; cout<<"\n"; } cout<<"\n"; }

#define UNIFORM_DOUBLE(a, b) (((b-a)*(double)rand()/RAND_MAX)+a) // [a, b) 

#define UNIFORM_LL(a, b) (ll)UNIFORM_DOUBLE(a, b) // [a, b) 

#define IN(v, lo, hi) ((lo)<=(v) && (v)<(hi))

#define DD(v) cout<<#v<<": "<<v<<endl

template <typename T0, typename T1> std::ostream& operator<<(std::ostream& os, const map<T0, T1>& v) { for( typename map<T0, T1>::const_iterator p = v.begin(); p!=v.end(); p++ ){os << p->first << ": " << p->second << " ";} return os; }

template <typename T0, typename T1> std::ostream& operator<<(std::ostream& os, const pair<T0, T1>& v) { os << v.first << ": " << v.second << " "; return os; }

template <typename T> std::ostream& operator<<(std::ostream& os, const vector<T>& v) { for( int i = 0; i < (int)v.size(); i++ ) { os << v[i] << " "; } return os; }

template <typename T> std::ostream& operator<<(std::ostream& os, const vector<vector<T> >& v) { for( int i = 0; i < (int)v.size(); i++ ) { os << v[i] << endl; } return os; }

template <typename T> std::ostream& operator<<(std::ostream& os, const set<T>& v) { vector<T> tmp(v.begin(), v.end()); os << tmp; return os; }

template <typename T> std::ostream& operator<<(std::ostream& os, const deque<T>& v) { vector<T> tmp(v.begin(), v.end()); os << tmp; return os; }



#define MOD 1000000007LL

#define INF (1LL<<60)



ll CS = 26;

//ll CS = 3;



ll N;

string s;



// nxt[c][i] ... i以降でcが最初に出てくる位置

VVI nxt;



using pll = pair<int, int>;



// i以降の文字列について, 最良の残りstepと選んだ文字

vector<pll> dp;



int main() {

	cin.tie(0);

	ios::sync_with_stdio(false);

	while(cin>>s) {

		N = s.size();

		dp = vector<pll>(N+1, MP(0, 0));

		// 範囲外なのでaでOK

		dp[N] = MP(1, 0);



		nxt = VVI(CS, VI(N+1, -1));

		for(int i=N-1;i>=0;i--) {

			REP(j, CS) if(s[i]-'a'==j) nxt[j][i] = i; else if(i+1<N) nxt[j][i] = nxt[j][i+1];

		}

		for(int i=N-1;i>=0;i--) {

			auto lans = MP(1<<30, 0);

			REP(j, CS) {

				int p = nxt[j][i];

				int l = 1;

				// p==-1のときは以降ないのでjでいい

				if(p!=-1) l = 1 + dp[p+1].first;

				if(l < lans.first) lans = MP(l, j);

			}

			dp[i] = lans;

		}

		auto vv = dp[0];

		string ans;

		ll p = 0;

		while(1) {

			int c = dp[p].second;

			p = nxt[c][p];

			ans.PB('a'+c);

			if(p==-1) break;

			p++;

		}

		cout<<ans<<endl;



//		break;

	}

	

	return 0;

}