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.

<image>
---
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.

<image>
---
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 JAVA):

Solution:
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class Main {
    static final int MAXT = 1000000;
    static final int depth = 30;
    static final double eps = 1e-4;

    static int n, m, l;
    static int[] b, h, f, a, p, t;
    static int[] width;
    static double[] height;

    static int toIndex(int pos) {
        int x = 0;
        for (int i = 0; i < n + 1; ++i) {
            x += width[i];
            if (x > pos) {
                return i;
            }
        }
        return -1;
    }

    // ind番目の枠にvolumeの水を入れる
    static void pour(int ind, double volume) {
        int minH = Math.min(h[ind], h[ind + 1]);

        // 枠の中にまだ水が入る
        if (height[ind] < minH) {
            // 水全てが入る
            if ((minH - height[ind]) * width[ind] * depth >= volume) {
                height[ind] += volume / (width[ind] * depth);
            } else {
                double tmp = minH - height[ind];
                height[ind] = minH;
                // 残りの水を同じ枠内に入れる
                pour(ind, volume - tmp * width[ind] * depth);
            }
        } else {
            // 水面が同じ高さの枠のリスト
            ArrayList<Integer> range = range(ind, height[ind]);
            int leftInd = range.get(0), rightInd = range.get(range.size() - 1);
            double nHeight = height[range.get(0)];

            // 左側に水が入る場合
            if (leftInd - 1 >= 0 && h[leftInd] <= nHeight && height[leftInd - 1] < nHeight) {
                pour(leftInd - 1, volume);
            } else if (rightInd + 1 <= n && h[rightInd + 1] <= nHeight && height[rightInd + 1] < nHeight) {
                pour(rightInd + 1, volume);
            } else {
                //　全体に水を入れる
                double retVolume = pourRange(range, volume);
                if (retVolume > eps ) {
                    pour(ind, retVolume);
                }
            }
        }
    }

    // rangeの枠全てに水を入れる　入りきらなかった水の量を返す
    static double pourRange(ArrayList<Integer> range, double volume) {
        int leftInd = range.get(0), rightInd = range.get(range.size() - 1);
        int minH = Math.min(h[leftInd], h[rightInd + 1]);
        int allWidth = 0;
        for (Integer i: range) {
            allWidth += width[i];
        }

        if ((minH - height[range.get(0)]) * allWidth * depth >= volume) {
            for (Integer i: range) {
                height[i] += volume / (allWidth * depth);
            }
            return 0;
        } else {
            double tmp = minH - height[range.get(0)];
            for (Integer i: range) {
                height[i] = minH;
            }
            return volume - tmp * allWidth * depth;
        }
    }

    // ind番目の枠と同じ高さの連続する枠のリストを返す(並び替え済み)
    static ArrayList<Integer> range(int ind, double target) {
        ArrayList<Integer> ret = new ArrayList<>();
        ret.add(ind);

        // 左側
        for (int i = ind - 1; i >= 0; --i) {
            if (h[i + 1] > target || height[i] != target) {
                break;
            }
            ret.add(i);
        }
        // 右側
        for (int i = ind + 1; i <= n; ++i) {
            if (h[i] > target || height[i] != target) {
                break;
            }
            ret.add(i);
        }
        Collections.sort(ret);
        return ret;
    }

    // pairじゃないけど許して
    static class Pair implements Comparable<Pair> {
        int first;
        int second;
        int third;
        Pair(int first, int second, int third) {
            this.first = first;
            this.second = second;
            this.third = third;
        }

        @Override
        public int compareTo(Pair o) {
            if (first == o.first) {
                return second - o.second;
            } else {
                return first - o.first;
            }
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int count = scanner.nextInt();

        for (int k = 0; k < count; ++k) {
            n = scanner.nextInt();
            b = new int[n + 2];
            h = new int[n + 2];
            height = new double[n + 1];
            width = new int[n + 1];

            b[0] = 0;
            b[n + 1] = 100;
            h[0] = 10000;
            h[n + 1] = 10000;

            for (int i = 1; i <= n; ++i) {
                b[i] = scanner.nextInt();
                h[i] = scanner.nextInt();

                width[i - 1] = b[i] - b[i - 1];
            }
            width[n] = b[n + 1] - b[n];

            m = scanner.nextInt();
            f = new int[m];
            a = new int[m];

            for (int i = 0; i < m; ++i) {
                f[i] = toIndex(scanner.nextInt());
                a[i] = scanner.nextInt();
            }

            l = scanner.nextInt();
            p = new int[l];
            t = new int[l];
            ArrayList<Pair> pairs = new ArrayList<>();

            for (int i = 0; i < l; ++i) {
                p[i] = toIndex(scanner.nextInt());
                t[i] = scanner.nextInt();
                pairs.add(new Pair(t[i], p[i], i));
            }

            Collections.sort(pairs);

            double[] ans = new double[l];

            for (int i = 0; i < l; ++i) {
                for (int j = 0; j < m; ++j) {
                    pour(f[j], a[j] * (pairs.get(i).first - (i == 0 ? 0 : pairs.get(i - 1).first)));
                }
                ans[pairs.get(i).third] = Math.min(50, height[pairs.get(i).second]);
            }

            for (int i = 0; i < l; ++i) {
                System.out.println(ans[i]);
            }
        }
    }
}