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: Time Limit Exceeded)

 #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
 #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
#include <unordered_set>
#include <iomanip>
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef vector<pair<int,int> > vpii;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<ll> > vvll;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define irep(it, stl) for(auto it = stl.begin(); it != stl.end(); it++)
#define drep(i,n) for(int i = (n) - 1; i >= 0; --i)
#define fin(ans) cout << (ans) << endl
#define mp(p,q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define Sort(a) sort(a.begin(), a.end())
#define Rort(a) sort(a.rbegin(), a.rend())
#define MATHPI acos(-1)
int dx8[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy8[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int dx4[4] = {1, 0, -1, 0};
int dy4[4] = {0, 1, 0, -1};
template <class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template <class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
template<typename A, size_t N, typename T> void Fill(A (&array)[N], const T &val){ fill((T*)array, (T*)(array+N), val); }
struct io{io(){ios::sync_with_stdio(false);cin.tie(0);}};
const int INF = INT_MAX;
const ll LLINF = 1LL<<60;
const ll MOD = 1000000007;
const double EPS = 1e-9;

//問題文が複雑な時は具体例を出して、小さい方から考えていく
//形だけが同じなら良くて、a~前の文字列で使った一番大きい文字+1が次の文字列で使える
//あとは形が同じでその中の辞書順最強をdfsで作っていく

int N;
string s = "";

void dfs(int v){
    //N文字になるまで
    if(v == N){
        fin(s);
    }
    
    
    //現在使える一番大きい文字
    char currentBigAlpha = 'a';
    
    //sからbiggestを探す
    while(s.find(currentBigAlpha) != string::npos){
        //aから初めてその文字列の一番大きいやつを探す
        currentBigAlpha++;
    }
    //次で+1してるので--しておく
    currentBigAlpha--;

    for(char c='a';c<=currentBigAlpha+1;c++){
        s += c;
        //全列挙のときdfsを使うのはここでN=3だったら最初にaaaまでいってもどったときにsをpopして次にaabをして
        //その次はab→aba→abb→abcとやるから、popが便利！
        dfs(s.size());
        s.pop_back();
    }

}
int main(){
      cin >> N;
      dfs(0);
      
}