Problem: Mr. Denjiro is a science teacher. Today he has just received a specially ordered water tank that will certainly be useful for his innovative experiments on water flow.

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Figure 1: The water tank

The size of the tank is 100cm (Width) * 50cm (Height) * 30cm (Depth) (see Figure 1). For the experiments, he fits several partition boards inside of the tank parallel to the sideboards. The width of each board is equal to the depth of the tank, i.e. 30cm. The height of each board is less than that of the tank, i.e. 50 cm, and differs one another. The boards are so thin that he can neglect the thickness in the experiments.

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Figure 2: The front view of the tank

The front view of the tank is shown in Figure 2.

There are several faucets above the tank and he turns them on at the beginning of his experiment. The tank is initially empty. Your mission is to write a computer program to simulate water flow in the tank.



Input

The input consists of multiple data sets. D is the number of the data sets.

D
DataSet1
DataSet2
...
DataSetD


The format of each data set (DataSetd , 1 <= d <= D) is as follows.

N
B1 H1
B2 H2
...
BN HN
M
F1 A1
F2 A2
...
FM AM
L
P1 T1
P2 T2
...
PL TL


Each line in the data set contains one or two integers.

N is the number of the boards he sets in the tank . Bi and Hi are the x-position (cm) and the height (cm) of the i-th board, where 1 <= i <= N .

Hi s differ from one another. You may assume the following.

> 0 < N < 10 ,
>  0 < B1 < B2 < ... < BN < 100 ,
>  0 < H1 < 50 , 0 < H2 < 50 , ..., 0 < HN < 50.

M is the number of the faucets above the tank . Fj and Aj are the x-position (cm) and the amount of water flow (cm3/second) of the j-th faucet , where 1 <= j <= M .

There is no faucet just above any boards . Namely, none of Fj is equal to Bi .

You may assume the following .

> 0 < M <10 ,
>  0 < F1 < F2 < ... < FM < 100 ,
>  0 < A1 < 100, 0 < A2 < 100, ... 0 < AM < 100.

L is the number of observation time and location. Pk is the x-position (cm) of the k-th observation point. Tk is the k-th observation time in seconds from the beginning.

None of Pk is equal to Bi .

You may assume the following .

> 0 < L < 10 ,
>  0 < P1 < 100, 0 < P2 < 100, ..., 0 < PL < 100 ,
>  0 < T1 < 1000000, 0 < T2 < 1000000, ... , 0 < TL < 1000000.

Output

For each data set, your program should output L lines each containing one real number which represents the height (cm) of the water level specified by the x-position Pk at the time Tk.

Each answer may not have an error greater than 0.001. As long as this condition is satisfied, you may output any number of digits after the decimal point.

After the water tank is filled to the brim, the water level at any Pk is equal to the height of the tank, that is, 50 cm.

Example

Input

2
5
15 40
35 20
50 45
70 30
80 10
3
20 3
60 2
65 2
6
40 4100
25 7500
10 18000
90 7000
25 15000
25 22000
5
15 40
35 20
50 45
70 30
80 10
2
60 4
75 1
3
60 6000
75 6000
85 6000


Output

0.666667
21.4286
36.6667
11.1111
40.0
50.0
30.0
13.3333
13.3333

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int,int>;
using ll = long long;
#define rep(i, j) for(int i=0; i < (int)(j); i++)
#define all(v) v.begin(),v.end()
const double EPS = 1e-8;

struct Node {
    int l, r, h;
    double water;
    bool is_child_full;
    list<Node> children; // all children shold have same height.
    Node(int l, int r, int h) : l(l), r(r), h(h), water(0), is_child_full(false) {}    
    void insert(int b, int h, int d = 0) {
        if(children.size() == 0) {
            children.push_back(Node(l, b, h));
            children.push_back(Node(b, r, h));
        } else {
            auto child = find(b);
            if(child->h != h) {
                child->insert(b, h, d+1);
            } else {
                int lc = child->l;
                child->l = b;
                children.insert(child, Node(lc, b, h));
            }
        }
    }
    // find child which range contains p
    list<Node>::iterator find(int p) {
        for(auto c = begin(children); c != end(children); c++) {
            if(c->l <= p and p < c->r) return c;
        }
        assert(0);
    }
    double capacity() {
        return 30.0 * h * (r - l);
    }
    double pour(int p, double a) {
        if(water + a >= capacity()) {
            double left = water + a - capacity();
            water = capacity();
            return left;
        }
        water += a;
        assert(water / 30.0 / (r - l) <= h);
        if(children.size() > 0) {
            auto child = find(p);
            double left = child->pour(p, a);
            if(left == 0) return 0;
            assert(abs(child->height(p) - child->h) < EPS); // child is full
            double left1 = left / 2, left2 = left / 2;
            // go right
            auto c2 = child;
            while(left1 > 0 and c2 != end(children)) {
                left1 = c2->pour(c2->l, left1);
                c2++;
            }
            // go left
            c2 = child;            
            left2 += left1;
            if(c2 != begin(children)) {
                do {
                    c2--;
                    left2 = c2->pour(c2->r - 1, left2);
                } while(left2 > 0 and c2 != begin(children));
            }
            // go right(2nd)
            c2 = child;
            while(left2 > 0 and c2 != end(children)) {
                left2 = c2->pour(c2->l, left2);
                c2++;
            }
            if(left2 > 0) is_child_full = true;
        }
        return 0;
    }
    double height(int p) {
        if(is_child_full or children.size() == 0) return water / 30.0 / (r - l);
        return find(p)->height(p);
    }
    void show_structure(int d = 0) {        
        if(is_child_full) return;
        rep(i, d) cerr << "\t";
        for(auto p : children) cerr << "[" << p.l << ", " << p.water / 30.0 / (p.r - p.l) << "/" << p.h << " , " << p.r << ")" << " ";
        cerr << endl;
        for(auto p : children) p.show_structure(d + 1);
    }
};


void solve() {
    int N; cin >> N;
    // make tree
    vector<pii> HB(N); rep(i, N) cin >> HB[i].second >> HB[i].first;
    sort(all(HB), greater<pii>());    
    Node root(0, 100, 50);
    rep(i, N) root.insert(HB[i].second, HB[i].first);
    //root.show_structure();
    
    // input pour
    int M; cin >> M;
    vector<int> F(M), A(M); rep(i, M) cin >> F[i] >> A[i];
    // input measure
    int L; cin >> L;
    vector<pair<pii,int>> TP(L);
    rep(i, L) {
        cin >> TP[i].first.second >> TP[i].first.first;
        TP[i].second = i;
    }
    sort(all(TP));

    // simulate for each t    
    int tp_i = 0;
    vector<double> ans(L, -1);
    rep(t, 1000000) {
        // measure point
        while(tp_i < L and TP[tp_i].first.first == t) {            
            double a = root.height(TP[tp_i].first.second);
            ans[TP[tp_i].second] = a;
            tp_i++;
        }
        // update height
        rep(m, M) root.pour(F[m], A[m]);
    }
    rep(i, L) printf("%.6lf\n", ans[i]);
}


int main() {
    int D; cin >> D;
    rep(i, D) solve();
    return 0;
}