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 <iostream>
#include <fstream>
#include <algorithm>
#include <vector>
#include <cstring>
//#define LOCAL
using namespace std;
typedef long long ll;
const int mx=1e5+10;
ll a[mx];

void display(ll n, ll num[mx]){
    for(ll i=0;i<n;i++){
        if(0==i) printf("%lld", num[i]);
        else printf(" %lld", num[i]);
    }
    puts("");
}

int main() {
#ifdef LOCAL
ifstream fin("1.txt");
streambuf*p;
p=cin.rdbuf(fin.rdbuf());
#endif // LOCAL
ll n;
cin>>n;
for(ll i=0;i<n;i++) cin>>a[i];
ll b[mx], c[mx];
memcpy(b, a, sizeof(a)), memcpy(c, a, sizeof(a));
if(prev_permutation(b, b+n)) display(n, b);
display(n, a);
if(next_permutation(c, c+n)) display(n, c);




#ifdef LOCAL
fin.close();
#endif // LOCAL
}