#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;struct Node {int num, vertex, weight, ansx, ansy;Node(int _n, int _v, int _w, int _ansx, int _ansy) : \num(_n), vertex(_v), weight(_w), ansx(_ansx), ansy(_ansy) {}bool operator < (const Node &other) const {return num < other.num;}
};const int N = 100005;
vector<Node>v[N], vv;
// first元素是询问的编号，pair的第一个元素是询问的
int ch[N][2], w[N];
int n, m;
int num[N<<1], cnt;
map<int,int>mp;
int lbit[N<<1], rbit[N<<1]; // 因为询问和给定节点加起来上限是2*N个不同的数值 

inline int lowbit(int x) {return x & -x;
}void add(int bit[], int x, int val) {for (int i = x; i <= cnt; i += lowbit(i)) {bit[i] += val;}
}int sum(int bit[], int x) {int ret = 0;for (int i = x; i > 0; i -= lowbit(i)) {ret += bit[i];}return ret;
}void dfs(int u) {for (int i = 0; i < (int)v[u].size(); ++i) {int weight = mp[v[u][i].weight];int lsum = sum(lbit, weight), rsum = sum(rbit, weight);int ltot = sum(lbit, cnt), rtot = sum(rbit, cnt);bool find = lsum && bool(lsum - sum(lbit, weight-1)) || rsum && bool(rsum - sum(rbit, weight-1));if (!find) {v[u][i].ansx = rsum;v[u][i].ansy = rtot-rsum + ltot-lsum + lsum*3 + rsum*3;} else {v[u][i].ansx = v[u][i].ansy = -1;}}if (ch[u][0]) {add(lbit, mp[w[u]], 1);dfs(ch[u][0]);add(lbit, mp[w[u]], -1);}if (ch[u][1]) {add(rbit, mp[w[u]], 1);dfs(ch[u][1]);add(rbit, mp[w[u]], -1);}
}int main() {int T;scanf("%d", &T);while (T--) {scanf("%d", &n);cnt = 0, mp.clear(), vv.clear();memset(ch, 0, sizeof (ch));memset(lbit, 0, sizeof (lbit));memset(rbit, 0, sizeof (rbit));for (int i = 1; i <= n; ++i) {v[i].clear();scanf("%d", &w[i]);num[cnt++] = w[i]; // 将所有的要进行处理的节点重量以及询问的重量离散化
        }scanf("%d", &m);int a, b, c;for (int i = 0; i < m; ++i) {scanf("%d %d %d", &a, &b, &c);ch[a][0] = b, ch[a][1] = c;}int Q;scanf("%d", &Q);for (int i = 0; i < Q; ++i) {scanf("%d %d", &a, &b);v[a].push_back(Node(i, a, b, 0, 0));num[cnt++] = b;}sort(num, num + cnt);cnt = unique(num, num + cnt) - num; // 离散化之后一共是cnt个元素 for (int i = 0; i < cnt; ++i) {mp[num[i]] = i + 1; // 这里加1是为了避免树状数组统计时无法处理0号元素
        }dfs(1); // 题目中约定了1为根for (int i = 1; i <= n; ++i) {for (int j = 0; j < (int)v[i].size(); ++j) {vv.push_back(v[i][j]);}}sort(vv.begin(), vv.end());for (int i = 0; i < (int)vv.size(); ++i) {printf(vv[i].ansx == -1 ? "0\n" : "%d %d\n", vv[i].ansx, vv[i].ansy);}}return 0;
}

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
struct Point {double x, y;Point() {}Point(double _x, double _y) : x(_x), y(_y) {}void read() {scanf("%lf %lf", &x, &y);}double operator * (const Point &b) const {return x*b.y - y*b.x;}Point operator - (const Point &b) const {return Point(x-b.x, y-b.y);}
};struct Line {Point s, e;Line() {}Line(Point _s, Point _e) : s(_s), e(_e) {}
};const int N = 105;
const int M = 105;
const double eps = 1e-6;
int n, m, p, LIM;
Point pt[N*3];
int seq[N];
double mp[N*3][N*3];
char G[N][N];
char vis[N];
int match[N];inline int sign(const double &x) {return (x < eps) ? -1 : (x > eps);
}inline double dist(const Point &a, const Point &b) {return sqrt(1.0*(a.x-b.x)*(a.x-b.x) + 1.0*(a.y-b.y)*(a.y-b.y));
}void floyd() {for (int k = 1; k <= LIM; ++k) {for (int i = 1; i <= LIM; ++i) {if (i == k) continue;for (int j = 1; j <= LIM; ++j)  {if (i == j || j == k) continue;if (sign(mp[i][k] + mp[k][j] - mp[i][j]) < 0) {mp[i][j] = mp[i][k] + mp[k][j];}}}}
}bool inter(const Line l1, const Line l2) {returnmax(l1.s.x, l1.e.x) > min(l2.s.x, l2.e.x) &&max(l2.s.x, l2.e.x) > min(l1.s.x, l1.e.x) &&max(l1.s.y, l1.e.y) > min(l2.s.y, l2.e.y) &&max(l2.s.y, l2.e.y) > min(l1.s.y, l1.e.y) &&sign((l2.s-l1.s)*(l1.e-l1.s)) * sign((l2.e-l1.s)*(l1.e-l1.s)) < 0 &&sign((l1.s-l2.s)*(l2.e-l2.s)) * sign((l1.e-l2.s)*(l2.e-l2.s)) < 0;
}void build() {bool c;for (int i = 1; i <= LIM; ++i) {for (int j = i+1; j <= LIM; ++j) { // 枚举两个城市之间是否有边相连 c = false;for (int k = 1, d = 1; k <= m; ++k, d+=2) {if (i > n && k == (i-n+1)/2) continue;if (j > n && k == (j-n+1)/2) continue;if (inter(Line(pt[i], pt[j]), Line(pt[n+d], pt[n+d+1]))) {c = true;break;}}if (!c) { // 说明没有直线与两个城市之间的连线相交 mp[i][j] = mp[j][i] = dist(pt[i], pt[j]);}}}floyd();
}bool path(int u) {for (int v = 1; v <= n; ++v) {if (vis[v] || !G[u][v]) continue;vis[v] = 1;if (!match[v] || path(match[v])) {match[v] = u;return true;}}return false;
}bool Ac(double mid) {memset(G, 0, sizeof (G));memset(match, 0, sizeof (match));for (int i = 1; i <= n; ++i) { // 有向无环图构造完成 for (int j = i+1; j <= n; ++j) {if (sign(mid - mp[seq[i]][seq[j]]) >= 0) {G[seq[i]][seq[j]] = 1;}}}int cnt = 0;for (int i = 1; i <= n; ++i) {memset(vis, 0, sizeof (vis));if (path(i)) ++cnt;}return n-cnt <= p;
}double bsearch(double l, double r) {double mid, ret;while (r - l >= eps) {mid = (l + r) / 2.0;if (Ac(mid)) {r = mid - eps;ret = mid;} else {l = mid + eps;}}return ret;
}int main() {int T;scanf("%d", &T);while (T--) {scanf("%d %d %d", &n, &m, &p);LIM = n+2*m;for (int i = 1; i <= LIM; ++i) {for (int j = i; j <= LIM; ++j) {mp[i][j] = mp[j][i] = 1e20;}}for (int i = 1; i <= n; ++i) pt[i].read();for (int i = 1, j = 1; i <= m; ++i, j+=2) {pt[n+j].read(), pt[n+j+1].read();}for (int i = 1; i <= n; ++i) scanf("%d", &seq[i]);build();double ret = bsearch(0, 1e5);printf("%.2f\n", ret);}return 0;
}