For this Question: Find the lexicographically smallest permutation P of integers from 1 to N such that:
1. It should not be strictly increasing i. e. there must be an i such that P[i] < P[i-1] for 2 <= i <= N
2. Sum of every three consecutive elements should be divisible by 3.The only line of input contains a single integer N, denoting the number of elements in the permutation.

3 <= N <= 15Print a single line containing N space separated integers denoting the elements of the permutation.Sample Input:
4

Sample Output:
1 3 2 4

Explanation:
Some permutations of 4 integers are:
1. [1, 2, 3, 4] - not valid since it is strictly increasing
2. [1, 3, 4, 2] - not valid since sum of [1, 3, 4] is not divisible by 3
3. [4, 3, 2, 1] - not valid because its not lexicographically smallest, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
 
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif

const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;

int a[N];

void solve(){
    int n;  cin >> n;
    for(int i = 1; i <= n; i++)
        a[i] = i;
    do{
        bool flag = 0;
        for(int i = 2; i <= n; i++)
            flag |= (a[i]-a[i-1] > 0 )?0:1;
        for(int i = 1; i <= n-2; i++){
            int sum = a[i] + a[i+1] + a[i+2];
            if(sum%3 != 0)
                flag = 0;
        }
        if(flag)
            break;
    }while(next_permutation(a+1, a+n+1));
    for(int i = 1; i <= n; i++)
        cout << a[i] << " ";
}

void testcases(){
    int tt = 1; 
    //cin >> tt;
    while(tt--){
        solve();
    }
}

signed main() {
    IOS;
        clock_t start = clock();

    testcases();

    cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms:";
    return 0;
}   , In this Programming Language: C++, Now tell me if this Code is compilable or not?