#include<iostream>
using namespace std;
const int N = 10, M = 7;
int e[N][N] = {0}, f[N], open[N];//e[i][j]表示i和j之间是否连通；f[i]表示结点i的父节点；open[i] 1表示结点i打开，0表示关闭 
long long ans = 0;int find(int x){if(f[x] == x) return x;return f[x] = find(f[x]); //压缩路径 
}void dfs(int n){if(n == 8){for(int i = 1; i <= M; i++) {f[i] = i;//并查集初始化，每个结点各为一个集合 }for(int i = 1; i <= M; i++){for(int j = 1; j <= M; j++){if(e[i][j] && open[i] && open[j]){//连通则合并 f[find(i)] = find(j);}}} int cnt = 0;//当前方案中的连通块有几个 for(int i = 1; i <= M; i++){if(open[i] && f[i] == i) cnt++;}	if(cnt == 1) ans++;//只有一个连通块，则为合法方案 return;}open[n] = 1;//打开二极管ndfs(n + 1);open[n] = 0;//关闭二极管n dfs(n + 1);
}int main(){e[1][2] = e[1][6] = 1;//把二极管视为顶点，是否连通视为之间是否有边 e[2][1] = e[2][3] = e[2][7] = 1;e[3][2] = e[3][7] = e[3][4] = 1;e[4][3] = e[4][5] = 1;e[5][4] = e[5][7] = e[5][6] = 1;e[6][1] = e[6][7] = e[6][5] = 1;e[7][2] = e[7][3] = e[7][6] = e[7][5] = 1;dfs(1); cout << ans;return 0;
}