1 #include<cstdio>
  2 #include<cstring>
  3 #include<vector>
  4 #include<iostream>
  5 #include<algorithm>
  6 using namespace std;
  7 
  8 const int N = 100010;
  9 
 10 int n;
 11 int a[9];
 12 int cntx, cnty;
 13 int x[N], y[N];
 14 int sum[N * 20], ls[N * 20], rs[N * 20], root[N], tot;
 15 vector<int> ins[N];
 16 
 17 struct Node
 18 {
 19     int x, y;
 20 }t[N];
 21 
 22 void update(int& q, int last, int l, int r, int v)
 23 {
 24     q = ++tot;
 25     ls[q] = ls[last];
 26     rs[q] = rs[last];
 27     sum[q] = sum[last] + 1;
 28 
 29     if(l == r) return;
 30     int mid = (l + r) >> 1;
 31     if (v <= mid) update(ls[q], ls[last], l, mid, v);
 32     else update(rs[q], rs[last], mid + 1, r, v);
 33 }
 34 
 35 int kth(int p, int q, int l, int r, int v) // 找第p版本线段树到第q版本线段树中差值为v的节点
 36 {
 37     if (l == r) return l;
 38 
 39     int s = sum[ls[q]] - sum[ls[p]];
 40     int mid = (l + r) >> 1;
 41     if (s >= v) return kth(ls[p], ls[q], l, mid, v);
 42     else return kth(rs[p], rs[q], mid + 1, r, v - s);
 43 }
 44 
 45 int query(int q, int l, int r, int L, int R) // 求区间和
 46 {
 47     if ((!q) || (L <= l && r <= R)) return sum[q];
 48 
 49     int res = 0;
 50     int mid = (l + r) >> 1;
 51     if (L <= mid) res += query(ls[q], l, mid, L, R);
 52     if (R > mid) res += query(rs[q], mid + 1, r, L, R);
 53     return res;
 54 }
 55 
 56 
 57 int S(int l1, int r1, int l2, int r2) // 求子矩阵 点[l1,l2] 到 点[r1,r2] 之间的点的数量
 58 {
 59     return query(root[r1], 1, cnty, l2, r2) - query(root[l1 - 1], 1, cnty, l2, r2);
 60 }
 61 
 62 bool solve() 
 63 {
 64     int l = 1, r = cntx, ans = 0; // 求在左边的那条竖线
 65     while (l <= r)
 66     {
 67         int mid = (l + r) >> 1;
 68         if (sum[root[mid]] >= a[0] + a[1] + a[2])
 69         {
 70             ans = mid;
 71             r = mid - 1;
 72         }
 73         else l = mid + 1;
 74     }
 75     if (ans == 0 || sum[root[ans]] != a[0] + a[1] + a[2]) return false; // 没找到这条线
 76     int p1 = ans;
 77 
 78     l = 1, r = cntx, ans = 0; // 求在右边的那条竖线
 79     while (l <= r)
 80     {
 81         int mid = (l + r) >> 1;
 82         if (sum[root[mid]] >= a[0] + a[1] + a[2] + a[3] + a[4] + a[5])
 83         {
 84             ans = mid;
 85             r = mid - 1;
 86         }
 87         else l = mid + 1;
 88     }
 89     if (ans == 0 || sum[root[ans]] != a[0] + a[1] + a[2] + a[3] + a[4] + a[5]) return false; // 没找到这条线
 90     int p2 = ans;
 91 
 92     int posx1 = kth(root[0], root[p1], 1, cnty, a[0]); // 找上面那条横线
 93     while (S(1, p1, 1, posx1) != a[0] || S(p1 + 1, p2, 1, posx1) != a[3] || S(p2 + 1, cntx, 1, posx1) != a[6])
 94     {
 95         if (S(1, p1, 1, posx1) != a[0]) return false; // 找不到
 96         ++posx1;
 97     }
 98 
 99     int posx2 = kth(root[0], root[p1], 1, cnty, a[0] + a[1]); // 找下面那条横线
100     while (S(1, p1, posx1+1, posx2) != a[1] || S(p1 + 1, p2, posx1 + 1, posx2) != a[4] || S(p2 + 1, cntx, posx1 + 1, posx2) != a[7])
101     {
102         if (S(1, p1, posx1 + 1, posx2) != a[1]) return false; // 找不到
103         ++posx2;
104     }
105 
106     printf("%.10lf %.10lf \n%.10lf %.10lf", x[p1] + 0.5, x[p2] + 0.5, y[posx1] + 0.5, y[posx2] + 0.5);
107     return true;
108 }
109 
110 bool com(Node a, Node b)
111 {
112     return a.x == b.x ? a.y < b.y : a.x < b.x;
113 }
114 
115 int main()
116 {
117     scanf("%d", &n);
118     for (int i = 1; i <= n; i++)
119     {
120         scanf("%d%d", &t[i].x, &t[i].y);
121         x[++cntx] = t[i].x;
122         y[++cnty] = t[i].y;
123     }
124     for (int i = 0; i < 9; i++) scanf("%d", &a[i]);
125 
126     sort(x + 1, x + 1 + cntx);
127     cntx = unique(x + 1, x + 1 + cntx) - (x + 1);
128     sort(y + 1, y + 1 + cnty);
129     cnty = unique(y + 1, y + 1 + cnty) - (y + 1);
130     sort(t + 1, t + 1 + n, com);
131 
132     for (int i = 1; i <= n; i++) // 哈希
133     {
134         t[i].x = lower_bound(x + 1, x + cntx + 1, t[i].x) - x;
135         t[i].y = lower_bound(y + 1, y + cnty + 1, t[i].y) - y;
136         ins[t[i].x].push_back(t[i].y);
137     }
138 
139     for (int i = 1; i <= cntx; i++) // 建可持久化线段树
140     {
141         root[i] = root[i - 1];
142         for (int j = 0; j < ins[i].size(); j++)
143         {
144             update(root[i], root[i], 1, cnty, ins[i][j]);
145         }
146     }
147 
148     int loop = 362880;
149     while (loop--)
150     {
151         next_permutation(a, a + 9); // 全排列
152         if (solve()) return 0;
153     }
154 
155     puts("-1");
156 }