Problem:
Score : 400 points

Problem Statement

In this problem, we only consider strings consisting of lowercase English letters.

Strings s and t are said to be isomorphic when the following conditions are satisfied:

|s| = |t| holds.

For every pair i, j, one of the following holds:

s_i = s_j and t_i = t_j.

s_i \neq s_j and t_i \neq t_j.

For example, abcac and zyxzx are isomorphic, while abcac and ppppp are not.

A string s is said to be in normal form when the following condition is satisfied:

For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison.

For example, abcac is in normal form, but zyxzx is not since it is isomorphic to abcac, which is lexicographically smaller than zyxzx.

You are given an integer N.
Print all strings of length N that are in normal form, in lexicographically ascending order.

Constraints

1 \leq N \leq 10

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Assume that there are K strings of length N that are in normal form: w_1, \ldots, w_K in lexicographical order.
Output should be in the following format:

w_1
:
w_K

Sample Input 1

1

Sample Output 1

a

Sample Input 2

2

Sample Output 2

aa
ab

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

#include "bits/stdc++.h"
#include <random>
#include <algorithm>
#include <unordered_set>
#define rep(i,n) for(int i = 0; i < n; i++)
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
#define vll vector<vector<long long>>
#define vl vector<long long>
#define vi vector<int>
#define vii vector<vector<int>>
#define pb push_back
#define pf push_front
#define ld long double
#define Sort(a) sort(a.begin(),a.end())
#define cSort(a,cmp) sort(a.begin(),a.end(),cmp)
#define reSort(a) sort(a.rbegin(), a.rend())
static const ll llMAX = numeric_limits<long long>::max();
static const int intMAX = numeric_limits<int>::max();
static const ll  llMIN = numeric_limits<long long>::min();
static const int intMIN = numeric_limits<int>::min();
static const ll d_5 = 100000;
static const ll d9_7 = 1000000007;
static const ll d_9 = 1000000000;
static const double PI=3.14159265358979323846;

template<class T>
void Printvector(std::vector<T> &a){
  int size = a.size();
  rep(i,size){
    cout<<a[i]<<" ";
  }
  cout<<endl;
}
template<class T>
void Printvector(std::vector<std::vector<T>> &a){
  int size = a.size();
  rep(i,size){
    int size2=a[i].size();
    rep(j,size2){
      cout<<a[i][j]<<" ";
    }
    cout<<endl;
  }
  cout<<endl;
}
template<class T>
vector<T> getaccum(vector<T> a){
  int size=a.size();
  vector<T> ans(size);
  ans[0]=a[0];
  for(int i=0;i<size-1;i++){
    ans[i+1]=ans[i]+a[i+1];
    //ans[i+1]%=d9_7;
  }
  return ans;
}
template<class T>
vector<vector<T>> getaccum2D(vector<vector<T>> a){
  int sizey=a.size();
  int sizex=a[0].size();
  vector<vector<T>> ans(sizey,vector<T>(sizex,0));
  //ans=a;
  ans[0][0]=a[0][0];
  for(int i=1;i<sizex;i++){
    ans[0][i]=ans[0][i-1]+a[0][i];
  }
  for(int i=1;i<sizex;i++){
    ans[i][0]=ans[i-1][0]+a[i][0];
  }
  for (int i = 0; i < sizey-1; i++)
        for (int j = 0; j < sizex-1; j++)
            ans[i+1][j+1] = ans[i][j+1] + ans[i+1][j] - ans[i][j] + a[i+1][j+1];
  return ans;
}
ll getaccumnum(vector<ll> accum,int l,int r){//閉区間[l,r]の総和
  if(l==0){
    return accum[r];
  }else{
    return accum[r]-accum[l-1];
  }
}
template<class T>
T getaccumnum2D(vector<vector<T>>& accum,int x1,int y1,int x2,int y2){//2の方が右下、閉区間
  
  T ans=accum[y2][x2];
  if(x1==0 && y1==0){
    ;
  }else if(x1>0 && y1==0){
    ans-=accum[y2][x1-1];
  }else if(x1==0 && y1>0){
    ans-=accum[y1-1][x2];
  }else{
    ans-=accum[y1-1][x2];
    ans-=accum[y2][x1-1];
    ans+=accum[y1-1][x1-1];
  }
  return ans;
}
ll digitpower(ll a,ll b){//aのb乗を計算
  if(b==1){
    return a;
  }else if(b==0){
    return 1;
  }
  int mode=0;
  if(b%2==1){
    ll tmp = digitpower(a,(b-1)/2);
    if(mode==1){
      tmp%=d9_7;
    }
    tmp*=tmp;
    if(mode==1){
      tmp%=d9_7;
    }
    tmp*=a;
    if(mode==1){
      return tmp%d9_7;
    }else{
      return tmp;
    }
  }else{
    ll tmp = digitpower(a,(b)/2);
    if(mode==1){
      tmp%=d9_7;
    }
    tmp*=tmp;
    if(mode==1){
      tmp%=d9_7;
    }
    if(mode==1){
      return tmp%d9_7;
    }else{
      return tmp;
    }
  }
}

vl facs(2000010,-1);
ll Factrial(ll num){
  if(facs[num]!=-1){
    return facs[num];
  }
  if(num==1||num<=0){
    return 1;
  }else if(num<0){
    printf("ERROR_minus\n");
    return 0;
  }else{
    facs[num]=(num*Factrial(num-1))%d9_7;
    return facs[num];
  }
}
long long modinv(long long a, long long m) {//modの逆元

    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;
}

/*vl invs(2000010,-1);
ll linercomb(ll n,ll k, ll mod){//n,kの線形時間で求める
if(n<k)return 0;
if(n<0)return 0;
if(k==0 || k==n)return 1;
  ll ans=Factrial(n);
  if(invs[k]==-1){
    invs[k]=modinv(Factrial(k),mod);
  }
  ans*=invs[k];
  ans%=d9_7;
  ll k1=Factrial(n-k);
  k1%=mod;
  ans*=modinv(k1,mod);
  ans%=mod;
  return ans;
}*/
unordered_map<ll,ll> prime_factor(int64_t n) {
  unordered_map<ll,ll> ret;
  for(int64_t i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}



struct datas{
  ll p;
  ll q;
};/*
bool cmp(const datas &a, const datas &b)
{
    return a.num < b.num;
}*/
template<class T>
T gcd(T a,T b){
  if(a==0){
    return b;

  }else if(b==0){
    return a;
  }
  while(1) {
    if(a < b) swap(a, b);
    if(!b) break;
    a %= b;
  }
  return a;
}
int LCS(string s,string t) {
  int n=s.size();
  int m=t.size();
  vector<vector<int>> dp=vector<vector<int>>(n+1,vector<int>(m+1,0));
    for (int i=0; i<n; ++i) {
        for (int j=0; j<m; ++j) {
            if (s[i] == t[j]) {
                dp[i+1][j+1] = dp[i][j] + 1;
            }
            else {
                dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j]);
            }
        }
    }
    return dp[n][m];
}

  int n,nn;
  set<string> ans; 
   void search(vi s,int nowmax){
     int flag=-1;
     rep(i,n){
       if(s[i]==-1){
         flag=i;
         break;
       }
     }
     if(flag==-1){
       string s3;
       rep(i,n){
         s3+=(char)(s[i]+'a');
       }
       ans.insert(s3);
       return;
     }
     nowmax++;
     s[flag]=nowmax;
     set<vi> did;
     int cnt=0;
    vi locs;
     rep(i,n){
       if(s[i]==-1){
         cnt++;
         locs.pb(i);
       }
     }
     rep(i,digitpower(2,cnt)){
       if(i==0)continue;
       vi s2=s;
       bitset<10> bits=i;
       rep(j,cnt){
         if(bits[j]==1){
           s2[locs[j]]=nowmax;
         }
       }
       search(s2,nowmax);
     }
       search(s,nowmax);
  }
int main(void){
  cin>>n;
  vi num(n,-1);
   nn=digitpower(2,n);
  search(num,-1);
  for(auto i:ans){
    cout<<i<<endl;
  }
  return 0;
}

//<<std::setprecision(30)

//重複削除
 /* std::sort(vec.begin(), vec.end());
 
  vec.erase(std::unique(vec.begin(), vec.end()), vec.end());*/

  //ペアの全探索
  /*do{
		int s=0;
		for(int i=0;i<n/2;i++)s+=v[i]^v[i+n/2];
		r=max(s,r);
		reverse(v.begin()+n/2,v.end());
	}while(next_permutation(v.begin(),v.end()));*/