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>
typedef long long ll;
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    bool test = false;
    if (prev_permutation(a.begin(), a.end())) {
        test = true;
        for (int j = 0; j < n; ++j) {
            if (j == 0) cout << a[j];
            else cout << " " << a[j];
        }
    }
    if (test) cout << endl;
    next_permutation(a.begin(), a.end());
    for (int j = 0; j < n; ++j) {
        if (j == 0) cout << a[j];
        else cout << " " << a[j];
    }
    cout << endl;
    test = false;
    if (next_permutation(a.begin(), a.end())) {
        test = true;
        for (int j = 0; j < n; ++j) {
            if (j == 0) cout << a[j];
            else cout << " " << a[j];
        }
    }
    if (test) cout << endl;
}