Problem: For given a sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$, print the previous permutation and the next permutation in lexicographic order.

Constraints

* $1 \leq n \leq 9$
* $a_i$ consist of $1, 2, ..., n$

Input

A sequence is given in the following format.


$n$
$a_0 \; a_1 \; ... \; a_{n-1}$


Output

Print the previous permutation, the given sequence and the next permutation in the 1st, 2nd and 3rd lines respectively. Separate adjacency elements by a space character. Note that if there is no permutation, print nothing in the corresponding line.

Examples

Input

3
2 1 3


Output

1 3 2
2 1 3
2 3 1


Input

3
3 2 1


Output

3 1 2
3 2 1

Solution (in CPP):

Solution:
#include <cstdio>
#include <utility>
#include <vector>
#include <algorithm>
using namespace std;

#define gcu getchar_unlocked
#define pcu putchar_unlocked
#define _il inline
#define _in _il int in
#define _sc _il bool scan
template <typename T>
T in(int c){T n=0;bool m=false;if(c=='-')m=true,c=gcu();
	do{n=10*n+(c-'0'),c=gcu();}while(c>='0');return m?-n:n;} //&&c<='9'
_in() {return in<int>(gcu());}
template <typename T>
T scan(T &n){int c=gcu();return c==EOF?false:(n=in<T>(c),true);}
_sc(char &c){c=gcu();gcu();return c!=EOF;}
//_sc(string &s){int c;s="";
//	for(;;){c=gcu();if(c=='\n'||c==' ')return true;else if(c==EOF)return false;s+=c;}}
template <typename H,typename... T> _sc(H &h, T&&... t){return scan(h)&&scan(t...);}
#define _vo _il void out
#define _vl _il void outl
template <typename T>
_vo(T n){static char buf[20];char *p=buf;
	if(n<0)pcu('-'),n*=-1;if(!n)*p++='0';else while(n)*p++=n%10+'0',n/=10;
	while (p!=buf)pcu(*--p);}
_vo(const char *s){while(*s)pcu(*s++);}
_vo(char c){pcu(c);}
//_vo(string &s){for(char c:s)pcu(c);}
//_vo(string s){for(char c:s)pcu(c);}
template <typename H,typename... T> _vo(H&& h, T&&... t){out(h);out(move(t)...);}
template <typename T> _vo(vector<T> &v){for(T &x:v)out(&x == &v[0]?"":" "),out(x);out('\n');}
_vl(){out('\n');}
template <typename... T> _vl(T&&... t){out(move(t)...);outl();}

int main() {
	vector<int> v(in());
	for (int &x: v)
		x = in();
	if(prev_permutation(v.begin(), v.end()))
		out(v);
	next_permutation(v.begin(), v.end());
	out(v);
	if(next_permutation(v.begin(), v.end()))
		out(v);
}