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<bits/stdc++.h>
using namespace std;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
#define rep(i, n) for(UL i = 0; i < (n); i++)

int main() {
 UL N;scanf("%u",&N);
 UL A[9]; rep(i,N) scanf("%u",&A[i]);
 UL pre[9],nx[9]; rep(i,N) pre[i]=nx[i]=A[i];
 for(UL i=N-1; i!=0; i--){
  if(pre[i-1]>pre[i]){
   for(UL j=i+1; j<N; j++) if(pre[i-1]>pre[j] && pre[i]<pre[j]) swap(pre[i],pre[j]);
   swap(pre[i-1],pre[i]);
   sort(pre+i,pre+N); reverse(pre+i,pre+N);
   rep(j,N){if(j)printf(" "); printf("%u",pre[j]);}
   printf("\n");
   break;
  }
 }
 rep(j,N){if(j)printf(" "); printf("%u",A[j]);}
 printf("\n");
 for(UL i=N-1; i!=0; i--){
  if(nx[i-1]<nx[i]){
   for(UL j=i+1; j<N; j++) if(nx[i-1]<nx[j] && nx[i]>nx[j]) swap(nx[i],nx[j]);
   swap(nx[i-1],nx[i]);
   sort(nx+i,nx+N);
   rep(j,N){if(j)printf(" "); printf("%u",nx[j]);}
   printf("\n");
   break;
  }
 }
 return 0;
}