#include <bits/stdc++.h>

using namespace std;
typedef long long LL;

int x[4] = {1, 2, 3, 4};
const int N = 110;
int a[N][4];
int size;

//填充一个全排列二维数组
void fill_qpl() {
    do {
        a[size][0] = x[0], a[size][1] = x[1], a[size][2] = x[2], a[size++][3] = x[3];
    } while (next_permutation(x, x + 4));
}

int bucket[5];

bool check(int a[4], int b[4], int c[4], int d[4]) {
    //左上
    memset(bucket, 0, sizeof bucket);
    bucket[a[0]]++, bucket[a[1]]++, bucket[b[0]]++, bucket[b[1]]++;
    for (int i = 1; i <= 4; i++)if (bucket[i] != 1) return false;
    //右上
    memset(bucket, 0, sizeof bucket);
    bucket[a[2]]++, bucket[a[3]]++, bucket[b[2]]++, bucket[b[3]]++;
    for (int i = 1; i <= 4; i++)if (bucket[i] != 1) return false;
    //左下
    memset(bucket, 0, sizeof bucket);
    bucket[c[0]]++, bucket[c[1]]++, bucket[d[0]]++, bucket[d[1]]++;
    for (int i = 1; i <= 4; i++)if (bucket[i] != 1) return false;
    //右下
    memset(bucket, 0, sizeof bucket);
    bucket[c[2]]++, bucket[c[3]]++, bucket[d[2]]++, bucket[d[3]]++;
    for (int i = 1; i <= 4; i++)if (bucket[i] != 1) return false;

    //竖着也需要检查噢~
    for (int k = 0; k < 4; k++) {
        memset(bucket, 0, sizeof bucket);
        bucket[a[k]]++, bucket[b[k]]++, bucket[c[k]]++, bucket[d[k]]++;
        for (int i = 1; i <= 4; i++)if (bucket[i] != 1) return false;
    }
    return true;
}

int cnt;

int main() {
    //填充
    fill_qpl();

    for (int i = 0; i < size; i++)
        for (int j = 0; j < size; j++)
            for (int k = 0; k < size; k++)
                for (int p = 0; p < size; p++) {
                    //四行数据
                    //a[i],a[j],a[k],a[p] 是否可以构成数独?
                    //构建的矩阵，切割成四块，每块用桶的思路看看是不是满足1，2，3，4分布
                    if (i == j || i == k || i == p || j == k || j == p || k == p) continue;

                    if (check(a[i], a[j], a[k], a[p])) {
                        //输出
                        for (int m = 0; m < 4; m++) cout << a[i][m] << " ";
                        cout << endl;
                        for (int m = 0; m < 4; m++) cout << a[j][m] << " ";
                        cout << endl;
                        for (int m = 0; m < 4; m++) cout << a[k][m] << " ";
                        cout << endl;
                        for (int m = 0; m < 4; m++) cout << a[p][m] << " ";
                        cout << endl;
                        cout << endl;

                        //计数
                        cnt++;
                    }
                }

    cout << cnt << endl;
    return 0;
}