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 <algorithm>
#include <stdio.h>
using namespace std;
int main(){
    bool big=true,less=true;
    int n,a[20],tot=1;
    scanf("%d%d",&n,&a[0]);
    for(int i=1;i<n;i++){
        scanf("%d",&a[i]);
        big&=a[i]<a[i-1];
        less&=a[i]>a[i-1];
        tot*=i;
    }
    tot*=n;
    for(int i=1;i<tot;i++)next_permutation(a,a+n);
    if(!less){
        printf("%d",a[0]);
        for(int i=1;i<n;i++)printf(" %d",a[i]);
        printf("\n");
    }
    next_permutation(a,a+n);
    printf("%d",a[0]);
    for(int i=1;i<n;i++)printf(" %d",a[i]);
    printf("\n");
    if(!big){
        next_permutation(a,a+n);
        printf("%d",a[0]);
        for(int i=1;i<n;i++)printf(" %d",a[i]);
        printf("\n");
    }
}