在一个小镇里面，有N户人家，每户人家的地址为（x，y，z）。现在要建立一个水循环系统，其中建立一个水井的费用与地址的高度成正比，即为X*z。而从一户有水井的人家接水到自己家的费用为：（|x2-x1| + |y2-y1| + |z2-z1|) * Y。当有水井的房子高度比其低时，需要加额外的Z元抽水费。

#include<iostream>
#include<fstream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 1100
#define INF 1<<25
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
typedef int type;
using namespace std;
int n, X, Y, Z, m;
struct point
{
    int x, y, z;
}p[maxn];
struct node
{
    int u, v;
    type w;
}e[maxn * maxn];
type dis(point a, point b)
{
    return abs(a.x - b.x) + abs(a.y - b.y) + abs(a.z - b.z);
}
int pre[maxn], id[maxn], vis[maxn];
type in[maxn];
type Directed_MST(int root, int NV, int NE)
{
    type ret = 0;
    while(true)
    {
        for (int i = 0; i < NV; i++) in[i] = INF;
        for (int i = 0; i < NE; i++)
        {
            int u = e[i].u, v = e[i].v;
            if (e[i].w < in[v] && u != v)
            {
                pre[v] = u;
                in[v] = e[i].w;
            }
        }
        for (int i = 0; i < NV; i++)
        {
            if (i == root) continue;
            if (in[i] == INF) return -1;
        }
        int cntcode = 0;
        mem(id, -1), mem(vis, -1);
        in[root] = 0;
        for (int i = 0; i < NV; i++)
        {
            ret += in[i];
            int v = i;
            while(vis[v] != i && id[v] == -1 && v != root)
            {
                vis[v] = i;
                v = pre[v];
            }
            if (v != root && id[v] == -1)
            {
                for (int u = pre[v]; u != v; u = pre[u])
                    id[u] = cntcode;
                id[v] = cntcode++;
            }
        }
        if (!cntcode) break;
        for (int i = 0; i < NV; i++) if (id[i] == -1)
            id[i] = cntcode++;
        for (int i = 0; i < NE; i++)
        {
            int v = e[i].v;
            e[i].u = id[e[i].u];
            e[i].v = id[e[i].v];
            if (e[i].u != e[i].v) e[i].w -= in[v];
        }
        NV = cntcode;
        root = id[root];
    }
    return ret;
}
int main ()
{
    while(scanf("%d%d%d%d", &n, &X, &Y, &Z) != EOF)
    {
        if (n + X + Y + Z == 0) break;
        m = 0;
        for (int i = 1; i <= n; i++)
        {
            scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].z);
            e[m].u = 0, e[m].v = i, e[m++].w = p[i].z * X;
        }
        for (int i = 1; i <= n; i++)
        {
            int k, d;
            scanf("%d", &k);
            while(k--)
            {
                scanf("%d", &d);
                e[m].u = i, e[m].v = d, e[m].w = dis(p[i], p[d]) * Y;
                if (p[i].z < p[d].z) e[m].w += Z;
                m++;
            }
        }
        type ans = Directed_MST(0, n + 1, m);
        if (ans == -1) puts("poor XiaoA");
        else printf("%d\n", ans);
    }
    return 0;
}