#include <bits/stdc++.h>
using namespace std;

using ull = unsigned long long;

int n;
ull ans = 0;

// 每行/每列已经出现过的形状、颜色
int rowShape[4], rowColor[4], colShape[4], colColor[4];

// usedPair[s] 的第 c 位为 1，表示 (shape=s, color=c) 已经用过
int usedPair[4];

void dfs(int rem) {
    if (rem == 0) {
        ++ans;
        return;
    }

    // 选择候选数最少的空格子（MRV剪枝）
    int bestIdx = -1;
    int bestCnt = 100;
    pair<int,int> bestCand[16];
    int bestLen = 0;

    int tmp = rem;
    while (tmp) {
        int idx = __builtin_ctz((unsigned)tmp);
        int i = idx / n, j = idx % n;

        pair<int,int> cand[16];
        int len = 0;

        for (int s = 0; s < n; ++s) {
            if ((rowShape[i] >> s) & 1) continue;
            if ((colShape[j] >> s) & 1) continue;
            for (int c = 0; c < n; ++c) {
                if ((rowColor[i] >> c) & 1) continue;
                if ((colColor[j] >> c) & 1) continue;
                if ((usedPair[s] >> c) & 1) continue;
                cand[len++] = {s, c};
            }
        }

        if (len == 0) return; // 这个格子已经无解，直接剪掉
        if (len < bestCnt) {
            bestCnt = len;
            bestIdx = idx;
            bestLen = len;
            for (int k = 0; k < len; ++k) bestCand[k] = cand[k];
            if (bestCnt == 1) break;
        }

        tmp &= tmp - 1;
    }

    int i = bestIdx / n, j = bestIdx % n;
    rem ^= (1 << bestIdx);

    for (int k = 0; k < bestLen; ++k) {
        int s = bestCand[k].first;
        int c = bestCand[k].second;

        rowShape[i] |= (1 << s);
        rowColor[i] |= (1 << c);
        colShape[j] |= (1 << s);
        colColor[j] |= (1 << c);
        usedPair[s] |= (1 << c);

        dfs(rem);

        usedPair[s] ^= (1 << c);
        colColor[j] ^= (1 << c);
        colShape[j] ^= (1 << s);
        rowColor[i] ^= (1 << c);
        rowShape[i] ^= (1 << s);
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n;

    if (n == 1) {
        cout << 1 << '\n';
        return 0;
    }
    if (n == 2) {
        cout << 0 << '\n';
        return 0;
    }

    int rem = (1 << (n * n)) - 1;
    dfs(rem);

    cout << ans << '\n';
    return 0;
}