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.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class C {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        C c = new C();
        while (t > 0){
            t--;
            c.solve(sc, pw);
        }
        pw.close();

    }
    class Node{
        int len;
        double prob;
        Node(int len, double prob){
            this.len = len;
            this.prob = prob;
        }
    }
     void solve(Scanner sc, PrintWriter pw){
        int n = sc.nextInt(), m = sc.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }
         List<Node> list = new ArrayList<>();
         for (int i = 0; i < m; i++) {
             list.add(new Node(sc.nextInt(), sc.nextDouble()));
         }
         int max = 0;
         int idx = 0;
         double res = 1;
         int[] le = new int[n];
         int[] ri = new int[n];
         for (int i = 0; i < n; i++) {
             max = Math.max(max, arr[i]);
             le[i] = max;
         }
         max = Integer.MAX_VALUE;
         for (int i = n - 1; i >= 0; i--) {
             max = Math.min(max, arr[i]);
             ri[i] = max;
         }
         int end = n;
         for (int i = n - 1; i >= 0; i--) {
             if(arr[i] == end){
                 end --;
             }else{
                 break;
             }
         }
         for(Node nd : list){
             if(nd.len >= end){
                 res *= (1 - nd.prob);
             }
         }
         if(end == 0){
             pw.println(1.0);
         }else{
             pw.println((1 - res));
         }
    }
}