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 java.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.valueOf(reader.readLine());
        while((t--) != 0) {
            HashMap<Integer, Double> map = new HashMap<>();
            
            String[] tmp = reader.readLine().split(" ");
            int n = Integer.valueOf(tmp[0]);
            int m = Integer.valueOf(tmp[1]);
            tmp = reader.readLine().split(" ");
            
            int[] info = new int[n];
            int[] sort = new int[n];
            for(int i=0; i<n; i++) {
                info[i] = Integer.valueOf(tmp[i]);
                sort[i] = info[i];
            }
            
            Arrays.sort(sort);
            
            for(int i=0; i<m; i++) {
                tmp = reader.readLine().split(" ");
                int p = Integer.valueOf(tmp[0]);
                if(map.get(p) != null) {
                    double q = map.get(p);
                    double w = Double.valueOf(tmp[1]);
                    map.put(p, 1 - (1-q)*(1-w));
                }else {
                    map.put(Integer.valueOf(tmp[0]), Double.valueOf(tmp[1]));
                }
            }
            
            double res = 0;
            if(map.get(n) != null) {
                res += map.get(n);
            }
            int i = n - 1;
            double notSort = 1 - res;
            for(i=n-1; i>=0; i--) {
                if(info[i] == sort[i]) {
                    res += notSort * (map.get(i)==null?0:map.get(i));
                    notSort *= (1 - (map.get(i)==null?0:map.get(i)));
                }else {
                    break;
                }
            }
            if(i == -1) {
                System.out.println(String.format("%.6f", 1.0).toString());
            }else {
                System.out.println(String.format("%.6f", res).toString());
            }
            
        }
    }
}