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;
using i64 = int64_t;
using vi = vector<i64>;
using vvi = vector<vi>;

void printvec(vi& x, int size = 0) {
    if (x.size() == 0) {
        cout << endl;
        return;
    }
    cout << x.front();
    for (int i = 1; i < (size == 0 ? x.size() : size); i++) {
        cout << " " << x[i];
    }
    cout << endl;
}

int main() {
    int n;
    cin >> n;
    vi as(n);
    for (int i = 0; i < n; i++) {
        cin >> as[i];
    }
    vi bs(as), cs(as);

    if (prev_permutation(bs.begin(), bs.end())) {
        printvec(bs);
    }
    printvec(as);
    if (next_permutation(cs.begin(), cs.end())) {
        printvec(cs);
    }
}