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 reps(i, n, m) for (int i = (int) (n); i < (int) (m); i++)
#define rep(i, n) reps(i, 0, (n))

int main() {
    int n; cin >> n;
    vector<int> v(n);
    rep(i, n) cin >> v[i];
    auto b = v;
    auto a = v;
    bool aa = prev_permutation(a.begin(), a.end());
    auto c = b;
    bool cc = next_permutation(c.begin(), c.end());
    if (aa) rep(i, n) cout << a[i] << " \n"[i==n-1];
    rep(i, n) cout << b[i] << " \n"[i==n-1];
    if (cc) rep(i, n) cout << c[i] << " \n"[i==n-1];
}