//给定 id id的父亲 id的母亲（-1表示过世） k id的k个孩子 房产的数量 房产的面积
//求每个家庭（用最小的id存储）房产的总数 平均房产数量和面积（平均保留三位）
//排序：平均面积的降序 tie id的增序
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct node0 {int id, fid, mid, num, area;int cid[10];
}data[1001];
struct node {int id, people;double num, area;bool flag = false;
}ans[10000];
vector<int> father(10000);//存放i的父节点的位置
vector<bool> visited(10000);
int find(int x){//找到x的根节点while (x != father[x])x = father[x];return x;
}
void Union(int a, int b){a = find(a);b = find(b);if(a > b) father[a] = b;else father[b] = a;
}
bool cmp(node &a, node &b){if(a.area == b.area) return a.id < b.id;else return a.area > b.area;
}int main(){int n, k, cnt = 0;cin >> n;for (int i = 0; i < 10000; i++)father[i] = i;//初始化每一个节点的父节点为本身for (int i = 0; i < n; i++) {cin >> data[i].id >> data[i].fid >> data[i].mid >> k;visited[data[i].id] = true;//记录存在该结点if (data[i].fid != -1) {visited[data[i].fid] = true;Union(data[i].fid, data[i].id);}if (data[i].mid != -1) {visited[data[i].mid] = true;Union(data[i].mid, data[i].id);}for (int j = 0; j < k; j++) {cin >> data[i].cid[j];visited[data[i].cid[j]] = true;Union(data[i].cid[j], data[i].id);}cin >> data[i].num >> data[i].area;}for (int i = 0; i < n; i++) {int id = find(data[i].id);//找祖先节点ans[id].id = id;//祖先节点为答案节点ans[id].num += data[i].num;//同一家族的房子数加和ans[id].area += data[i].area;ans[id].flag = true;//记录存在该答案结点}for (int i = 0; i < 10000; i++) {if(visited[i]) ans[find(i)].people++;if(ans[i].flag) cnt++;}for (int i = 0; i < 10000; i++) {if(ans[i].flag){ans[i].num = ans[i].num * 1.0 / ans[i].people;ans[i].area = ans[i].area * 1.0 / ans[i].people;}}sort(ans, ans + 10000, cmp);cout << cnt << endl;for (int i = 0; i < cnt; i++) {printf("%04d %d %.3lf %.3lf\n", ans[i].id, ans[i].people, ans[i].num, ans[i].area);}return 0;
}