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 <cstdio>
#include <vector>
#include <algorithm>
#include <map>
#include <numeric>
#include <string>
#include <cmath>
#include <iomanip>
#include <queue>
#include <list>
#include <stack>
#include <cctype>
#include <cmath>

using namespace std;

/* typedef */

typedef long long ll;

/* class */
/* constant */

const int INF = 1 << 30;
const int MAX = 100000;
const int mod = 1000000007;
const double pi = 3.141592653589;
/* global variables */

/* function */
void printVec(vector<int> &vec);
/* main */
int main() {

    int N; cin >> N;
    vector<int> v(N);
    for (int i = 0; i < N; i++) cin >> v[i];
    vector<int> v_prev(v.begin(), v.end());
    vector<int> v_next(v.begin(), v.end());

    int can_pre = prev_permutation(v_prev.begin(), v_prev.end());
    int can_next = next_permutation(v_next.begin(), v_next.end());

    if (can_pre) 
        printVec(v_prev);
    printVec(v);
    if (can_next)
        printVec(v_next);
}

void printVec(vector<int> &vec) {

    for (int i = 0; i < vec.size(); i++) {
        if (i) cout << ' ';
        cout << vec[i];
    } cout << '\n';
}