Global Round 27 C

#include <bits/stdc++.h>
using namespace std;
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());
#define NeedForSpeed                  \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
#define f first
#define s second
#define endl "\n"
const int mod = 1000000007;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
void solve()
{
    // for n -> odd, we are bounded as last is &, so we keep (n-1, n) at the last as (n-1 & n) = n-1
    // but we can go more than (n-1), by turing on the last bit of (n-1) (which is even) so we use (3,1)
    // for n -> even, try to chase 2^k -1
    int n;
    cin >> n;
    if (n % 2 == 1)
    {
        cout << n << endl;
        // odd case
        for (int i = 1; i <= n - 2; i++)
        {
            if (i != 1 && i != 3)
            {
                cout << i << " ";
            }
        }
        cout << 3 << " " << 1 << " " << (n - 1) << " " << n << endl;
    }
    else
    {
        // if it is a power of 2, then we can keep the last as n and n-1 before it to reach 2^k-1
        int lg = 0;
        while ((1 << lg) <= n)
        {
            lg++;
        }
        lg--;
        cout << ((1 << (lg + 1)) - 1) << endl;
        int m = 0;
        vi a(n + 1);
        for (int i = 1; i <= n; i++)
        {
            a[i] = i;
            m |= i;
        }
        int c = 1;
        int r = log2(m);
        for (int i = 0; i < r; i++)
        {
            c *= 2;
        }
        // not very intutive
        swap(a[n], a[c]);
        swap(a[c - 1], a[n - 1]);
        swap(a[c - 2], a[n - 2]);
        swap(a[1], a[n - 4]);
        swap(a[n - 3], a[3]);
        for (int i = 1; i <= n; i++)
        {
            cout << a[i] << " ";
        }
        cout << endl;
    }
    return;
}

signed main()
{
    NeedForSpeed;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}