Problem:
Score : 300 points

Problem Statement

Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.

A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, atcoder, zscoder and agc are diverse words while gotou and connect aren't diverse words.

Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.

Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.

Constraints

1 \leq |S| \leq 26

S is a diverse word.

Input

Input is given from Standard Input in the following format:

S

Output

Print the next word that appears after S in the dictionary, or -1 if it doesn't exist.

Sample Input 1

atcoder

Sample Output 1

atcoderb

atcoderb is the lexicographically smallest diverse word that is lexicographically larger than atcoder. Note that atcoderb is lexicographically smaller than b.

Sample Input 2

abc

Sample Output 2

abcd

Sample Input 3

zyxwvutsrqponmlkjihgfedcba

Sample Output 3

-1

This is the lexicographically largest diverse word, so the answer is -1.

Sample Input 4

abcdefghijklmnopqrstuvwzyx

Sample Output 4

abcdefghijklmnopqrstuvx

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

#include <bits/stdc++.h>
#include <stdio.h>
#include<iostream>
#include<cstdio>
#include<bitset>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<string.h>
#include<cmath>
#include<utility>
#include<functional>
#include<map>
#include<set>
#include<cctype>
#include<fstream>
#include <numeric>
#include <iomanip>
#include <cstring> 
using namespace std;
using ll=long long;


#define FOR(i, a, b) for( long long int i=(a);i<=(b);i++)
#define RFOR(i, a, b) for(long long int i=(a);i>=(b);i--)
#define MOD 1000000007
#define LLONG_MAXs 9223372036854775800

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

#include <iostream>
#include <cmath>
using namespace std;
 

 
bool isPrimeNum( ll x ){ // 素数である場合 true を返す
  if( x <= 1 ){ // 1以下である場合は素数でないことがすぐにわかる
    return false;
  }
  // sqrt( double型 ) は引数の平方根を double型で返すので、int型でキャスト
  int n = (int)sqrt( (double)x );
  for( int i = 2; i <= n; i++ ){
    if( x % i == 0 ){ // 割り切る整数がある場合、即判定終了
      return false;
    }
  }
  return true; // 割り切る整数がない場合、素数である
}

ll myPow(ll x, ll n, ll m){
  if(n == 0)
    return 1;
  if(n % 2 == 0)
    return myPow(x * x % m, n / 2, m);
  else
    return x * myPow(x, n - 1, m) % m;
}

constexpr ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
constexpr ll lcm(ll a,ll b){return a*b/gcd(a,b);}
constexpr ll abs(ll a,ll b){
if(a>=b)return a-b;
if(a<b)return b-a;
}
constexpr double dabs(double a,double b){
if(a>=b)return a-b;
if(a<b)return b-a;
}
constexpr ll min(ll a,ll b){
if(a>=b)return b;
if(a<b)return a;
}

constexpr ll max(ll a,ll b){
if(a>=b)return a;
if(a<b)return b;
}

int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

class UnionFind {
public:
	//親の番号を格納する。親だった場合は-(その集合のサイズ)
	vector<int> Parent;
 
	//作るときはParentの値を全て-1にする
	//こうすると全てバラバラになる
	UnionFind(int N) {
		Parent = vector<int>(N, -1);
	}
 
	//Aがどのグループに属しているか調べる
	int root(int A) {
		if (Parent[A] < 0) return A;
		return Parent[A] = root(Parent[A]);
	}
 
	//自分のいるグループの頂点数を調べる
	int size(int A) {
		return -Parent[root(A)];//親をとってきたい]
	}
 
	//AとBをくっ付ける
	bool connect(int A, int B) {
		//AとBを直接つなぐのではなく、root(A)にroot(B)をくっつける
		A = root(A);
		B = root(B);
		if (A == B) {
			//すでにくっついてるからくっ付けない
			return false;
		}
 
		//大きい方(A)に小さいほう(B)をくっ付けたい
		//大小が逆だったらひっくり返しちゃう。
		if (size(A) < size(B)) swap(A, B);
 
		//Aのサイズを更新する
		Parent[A] += Parent[B];
		//Bの親をAに変更する
		Parent[B] = A;
 
		return true;
	}
};




long long fac[510000], finv[510000], inv[510000];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < 510000; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
string replaceAll(string &replacedStr, string from, string to) {
    unsigned int pos = replacedStr.find(from);
    int toLen = to.length();
 
    if (from.empty()) {
        return replacedStr;
    }
 
    while ((pos = replacedStr.find(from, pos)) != std::string::npos) {
        replacedStr.replace(pos, from.length(), to);
        pos += toLen;
    }
    return replacedStr;
}

void yn(bool flag) {
    if(flag){
      cout<<"Yes"<<endl;
    }else{
      cout<<"No"<<endl;
    }
    return;
}
void YN(bool flag) {
    if(flag){
      cout<<"YES"<<endl;
    }else{
      cout<<"NO"<<endl;
    }
    return;
}

struct Edge {
    int to;     // 辺の行き先
    int weight; // 辺の重み
    Edge(int t, int w) : to(t), weight(w) { }
};
//using Graph = vector<vector<Edge>>;
using Graph = vector<vector<int>>;
 
vector<bool> seen;
vector<ll> step;
void dfs(const Graph &G, int v , ll now) {
    seen[v] = true; // v を訪問済にする
    step[v] = now;
    // v から行ける各頂点 next_v について
    for (auto next_v : G[v]) { 
        if (seen[next_v] && (now+1 >= step[next_v]) ) continue; // next_v が探索済だったらスルー
        dfs(G, next_v, now+1); // 再帰的に探索
    }
}
std::vector<int> enum_div(int n)//nの約数を列挙
{
	std::vector<int> ret;
	for(int i=1 ; i*i<=n ; ++i)
	{
		if(n%i == 0)
		{
			ret.push_back(i);
			if(i!=1 && i*i!=n)
			{
				ret.push_back(n/i);
			}
		}
	}
  ret.push_back(n);
	return ret;
}

void Ssort(int no, char *month[]){
    int i, j;
    char *temp;
    for(i=0; i<no; i++){
        for(j=i+1; j<no; j++){
            if(strcmp((month[i]),(month[j]))>0){
                temp = *(month+i);
                *(month+i) = *(month+j);
                *(month+j) = temp;
            }
        }    
    }
}

int main() {
  string S;
  cin>>S;
  if(S=="zyxwvutsrqponmlkjihgfedcba"){
    cout<<"-1"<<endl;
    return 0;
  }
  if(S.size()==26){

    RFOR(i,S.size()-1,1){
      if(S[i-1]<S[i]){

        FOR(j,0,i-2){
          char c = (char)S[j];
          cout<<c;
        }
      FOR(ii,S[i-1]+1,'z'){
        ll flag=0;
          FOR(j,0,i-1){
            if(ii==S[j])flag++;
          }

        if(flag==0){
            cout<<(char)ii<<endl;
            return 0;
        }
        
      }


      }
    }

  }else{
    FOR(i,'a','z'){

      ll flag=0;
      FOR(j,0,S.size()-1){
        if(i==S[j])flag++;
      }
      if(flag==0){
        char c = (char)i;
        cout<<S<<c<<endl;
        return 0;
      }
    }
  }

}