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;
#define ll long long
#define r(i, n) for(int i=0;i<n;i++)

int main() {
    ll n;
    cin >> n;
    vector<ll> v(n);
    r(i, n)cin >> v[i];
    if (prev_permutation(v.begin(), v.end())) {
        r(i, n) {
            cout << v[i] << " \n"[i == n - 1];
        }
    }
    next_permutation(v.begin(), v.end());
    r(i, n) {
        cout << v[i] << " \n"[i == n - 1];
    }
    if (next_permutation(v.begin(), v.end())) {
        r(i, n) {
            cout << v[i] << " \n"[i == n - 1];
        }
    }

    return 0;
}