Problem: Ron is a happy owner of a permutation a of length n.

A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

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Ron's permutation is subjected to m experiments of the following type: (r_i, p_i). This means that elements in range [1, r_i] (in other words, the prefix of length r_i) have to be sorted in ascending order with the probability of p_i. All experiments are performed in the same order in which they are specified in the input data.

As an example, let's take a look at a permutation [4, 2, 1, 5, 3] and an experiment (3, 0.6). After such an experiment with the probability of 60\% the permutation will assume the form [1, 2, 4, 5, 3] and with a 40\% probability it will remain unchanged.

You have to determine the probability of the permutation becoming completely sorted in ascending order after m experiments.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).

The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 10^5) — the length of the permutation and the number of experiments, respectively.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — contents of the permutation.

The following m lines of each test case each contain an integer r_i and a real number p_i (1 ≤ r_i ≤ n, 0 ≤ p_i ≤ 1) — the length of the prefix and the probability of it being sorted. All probabilities are given with at most 6 decimal places.

It is guaranteed that the sum of n and the sum of m does not exceed 10^5 (∑ n, ∑ m ≤ 10^5).

Output

For each test case, print a single number — the probability that after all experiments the permutation becomes sorted in ascending order. Your answer will be considered correct if its absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Example

Input


4
4 3
4 3 2 1
1 0.3
3 1
4 0.6
5 3
4 2 1 3 5
3 0.8
4 0.6
5 0.3
6 5
1 3 2 4 5 6
4 0.9
5 0.3
2 0.4
6 0.7
3 0.5
4 2
1 2 3 4
2 0.5
4 0.1


Output


0.600000
0.720000
0.989500
1.000000

Note

Explanation of the first test case: It can be demonstrated that whether the final permutation is sorted or not depends solely on sorting being performed in the (4, 0.6) experiment.

Solution (in JAVA):

Solution:
import javax.naming.TimeLimitExceededException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader bi = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(bi.readLine());
        for (int t = 0; t < T; t++) {
            String[] nm = bi.readLine().split("\\s");
            int n = Integer.parseInt(nm[0]), m = Integer.parseInt(nm[1]);
            int[] array;
            int wrongPosition = -1;
            array = Arrays.stream(bi.readLine().split("\\s")).mapToInt(Integer::parseInt).toArray();
            for (int i = 0; i < n; i++) {
                wrongPosition = (array[i] != i + 1 ? i : wrongPosition);
            }

            ArrayList<Integer> postions = new ArrayList<>();
            ArrayList<Integer> rs = new ArrayList<>();
            ArrayList<Double> ps = new ArrayList<>();
            for (int i = 0; i < m; i++) {
                String[] rp = bi.readLine().split("\\s");
                rs.add(Integer.parseInt(rp[0]) - 1);
                ps.add(Double.parseDouble(rp[1]));
                if (rs.get(i) >= wrongPosition) {
                    postions.add(i);
                }
            }

            if (wrongPosition == -1) {
                System.out.println(1.0);
                continue;
            }
            int amPositions = postions.size();
            if (amPositions == 1) {
                System.out.println(ps.get(postions.get(0)));
                continue;
            }
            if (amPositions == 0) {
                System.out.println(0);
                continue;
            }

            double[][] result = new double[amPositions][2];
            result[0][0] = 1 - ps.get(postions.get(0));
            result[0][1] = ps.get(postions.get(0));
            for (int i = 1; i < amPositions; i++) {
                double currP = ps.get(postions.get(i));
                result[i][0] = (1 - currP) * result[i - 1][0] + (1 - currP) * result[i - 1][1];
                result[i][1] = currP * result[i - 1][0] + currP * result[i - 1][1];
            }
            double allNotIncluded = 1;
            for (int i = 0; i < amPositions; i++) {
                allNotIncluded *= 1 - ps.get(postions.get(i));
            }
            System.out.println(result[amPositions - 1][0] + result[amPositions - 1][1] - allNotIncluded);
        }
    }
}