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>
#include <cmath>

using namespace std;
typedef long long ll;
//typedef unsigned long long ll;

#define rep(i, n) for (int i = 0; i < (n); ++i)
//#define rep(i, n) for (ll i = 0; i < (n); ++i)
//#define sz(x) ll(x.size())
//typedef pair<ll, int> P;
typedef pair<ll, ll> P;
//const double INF = 1e10;
const ll INF = LONG_LONG_MAX / 10;
//const ll INF = 1e15;
const ll MINF = LONG_LONG_MIN;
//const int INF = INT_MAX / 10;
#define cmin(x, y) x = min(x, y)
#define cmax(x, y) x = max(x, y)
//typedef pair<int, int> P;
//typedef pair<double, double> P;

bool contain(set<P> &s, P a) { return s.find(a) != s.end(); }

//ifstream myfile("C:\\Users\\riku\\Downloads\\0_00.txt");
//ofstream outfile("log.txt");
//outfile << setw(6) << setfill('0') << prefecture << setw(6) << setfill('0') << rank << endl;
// std::cout << std::bitset<8>(9);


void print_line(vector<ll> &line) {
    if (line.size() == 0ll) {
        cout << endl;
        return;
    }
    for (ll i = 0; i < line.size(); i++) {
        cout << line[i];
        if (i == line.size() - 1) cout << endl;
        else cout << ' ';
    }
}


typedef priority_queue<long long, vector<long long>, greater<long long>> PQ_ASK;
const int mod = 1000000007;

int main() {
    int n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n) cin >> a[i];

    if (prev_permutation(a.begin(), a.end()))
        print_line(a);

    next_permutation(a.begin(), a.end());
    print_line(a);

    if (next_permutation(a.begin(), a.end()))
        print_line(a);


}