Problem: Problem

Den, the phone number of Ukunikia Co., Ltd., enters a very long phone number into the phone every day.
One day, too tired, Den came up with a surprising idea.
"Isn't it even a little easier if you rearrange the arrangement of the buttons on the phone ?!"

The phone has squares evenly spaced at $ 3 \ times 3 $, and each of the nine squares has one button from 1 to 9 that can be sorted.
When typing a phone number, Den can do two things with just one hand:


* Move your index finger to touch one of the adjacent buttons on the side of the button you are currently touching.
* Press the button that your index finger is touching.



Initially, the index finger can be placed to touch any of the buttons 1-9.
Mr. Den thinks that the arrangement that can minimize the number of movements of the index finger from pressing the first button to the end of pressing the last button is efficient.

Now, here is the phone number of the customer with a length of $ N $.
What kind of arrangement is most efficient when considering only the customer's phone number?
Make the arrangement by rearranging the buttons.

Constraints

The input satisfies the following conditions.

* $ 1 \ leq N \ leq 10 ^ 5 $
* $ S $ is a string consisting of any number from 1 to 9.

Input

The input is given in the following format.


$ N $
$ S $


The first line gives the customer's phone number length $ N $.
The customer's phone number is given to the first line on the second line.

Output

Output the most efficient placement with no blanks on the 3 lines.
However, if there are multiple possible answers, start from the upper left frame.
one two Three
456
789
When arranging the numbers in the order of, output the one that is the smallest in the dictionary order.

Examples

Input

10
1236547896


Output

123
456
789


Input

11
31415926535


Output

137
456
892

Solution (in CPP):

Solution:
/////////////////////////////////////////////////////////
//
//  c/temp.cpp file
//  Last Updated: 2018-10-23 ...Maybe
//
//    I hope you adding this code to the setting file
//    alias g++='g++ -std=c++1y -DDEBUG_LOCAL'
//
/////////////////////////////////////////////////////////

#include <bits/stdc++.h>

using namespace std;

#define fi first
#define se second
#define mp make_pair
#define pb push_back

//#define int long long

#define _overload3(_1,_2,_3,name,...) name
#define _REP(i,n) REAP(i,0,n)
#define REAP(i,a,b) for(int i=int(a);i<int(b);++i)
#define REP(...) _overload3(__VA_ARGS__,REAP,_REP,)(__VA_ARGS__)

#define _REPR(i,n) REAPR(i,n,0)
#define REAPR(i,a,b) for(int i=int(a-1);i>=int(b);--i)
#define REPR(...) _overload3(__VA_ARGS__,REAPR,_REPR,)(__VA_ARGS__)

#define ALL(a) a.begin(),a.end()
#define rALL(a) a.rbegin(),a.rend()

#define coutALL(a) {int loop_coutALL=0;for(auto e:a) cout<<(loop_coutALL++?" ":"")<<e; cout<<endl;}

#define coutYN(a) cout<<((a)?"YES":"NO")<<endl;
#define coutYn(a) cout<<((a)?"Yes":"No")<<endl;
#define coutyn(a) cout<<((a)?"yes":"no")<<endl;

#define pcnt __builtin_popcount
#define buli(x) __builtin_popcountll(x)

const int INF=1145141919;
//const long long INF=114514191911451419;
const int MOD=(int)1e9+7;
const double EPS=1e-12;
const int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
//const int dx[]={-1,-1,-1,0,1,1,1,0},dy[]={-1,0,1,1,1,0,-1,-1};

using vi=vector<int>;
using vvi=vector<vi>;
using vs=vector<string>;

typedef pair<int,int> pii;
typedef pair<int,string> pis;
typedef pair<string,int> psi;
typedef pair<string,string> pss;

typedef long long ll;

template<typename T> istream &operator >> (istream &is, vector<T> &v){for(T &x:v) is>>x;return is;}
template<typename T> ostream &operator << (ostream &os, vector<T> &v){int i=0;for(T &x:v) os<<(i++?" ":"")<<x;return os;}

ll rev(ll n){ll x=0;for(;n>0;){x=x*10+n%10;n/=10;}return x;}

ll upper(ll n,ll m){return (n+m-1)/m;};
ll rounding(ll n){return (long double)n+0.5;};

bool inG(ll x,ll n){return 0<=x && x<n;}
bool outG(ll x,ll n){return x<0 || n<=x;}

inline int qp(int a,ll b){if(!b)return 1;int ans=1;do{if(b&1)ans=1ll*ans*a;a=1ll*a*a;}while(b>>=1);return ans;}
inline int qp(int a,ll b,int mo){if(!b)return 1;int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}

inline ll fac(ll k,ll n){ll a=1;for(int i=0;i<n;i++)a*=k--;return a;}
inline ll fac(ll k,ll n,int mo){ll a=1;for(int i=0;i<n;i++)a*=k--,a%=mo;return a;}

inline int dsum(ll n){int a=0;for(;n;n/=10)a+=n%10;return a;}

struct Arithmetic{Arithmetic(){cin.tie(0);ios::sync_with_stdio(0);cout<<fixed<<setprecision(20);}};

//#define DEBUG_LOCAL

#ifdef DEBUG_LOCAL
template<typename T> void deb(T a){cerr<<"deb: "<<a<<"です"<<endl;}
#define debl {cerr<<"debug: "<<__LINE__<<"行目だ"<<endl;}
void what_cr(){cout<<__GCC_ATOMIC_CHAR16_T_LOCK_FREE<<" ←？"<<endl;}
//ori_func S

//ori_func E
#else
template<typename T>void deb(T a){}
#define debl
void what_cr(){}void t_t(){}
#endif

signed main(){

  Arithmetic Exception;

  int n;

  cin>>n;

  string s;
  cin>>s;

  n=s.size();

  int how[9][9]={};

  REP(i,n-1) how[s[i]-'1'][s[i+1]-'1']++;

  vi a(9);

  REP(i,9) a[i]=i;

  vector <pair <int,vi> > ans;

  do{

    pii li[9];

    REP(i,9){

      li[a[i]].fi=i/3;
      li[a[i]].se=i%3;

    }

    int tmp=0;

    REP(i,9) REP(j,9) tmp += how[i][j]*(abs(li[i].fi-li[j].fi)+abs(li[i].se-li[j].se));

    ans.pb(mp(tmp,a));

  } while(next_permutation(ALL(a)));

  sort(ALL(ans));

  REP(i,9) cout<<ans[0].se[i]+1<<(i%3==2?"\n":"");

  return 0;

}