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.*;
import java.util.*;
public class Main {

	public static final int mod=1000000007;	
	static long[][] dp;
	static Map<Integer,Integer> mp;
	static Set<Integer> set;
	public static void main(String[] args) throws IOException {
		
		//File filein = new File("input.txt");
		//BufferedReader sc=new BufferedReader(new FileReader(filein));
		//PrintWriter out=new PrintWriter(new FileWriter("output.txt"));		

		//Scanner sc=new Scanner(System.in);
		BufferedReader sc=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out=new PrintWriter(System.out);
		int tt=Integer.parseInt(sc.readLine());
		int t=tt;
		mp=new HashMap<>();
		set=new HashSet<>();
		while(t-->0){
			out.flush();
			set.clear();
			String[] ip=sc.readLine().split(" ");
			int n=Integer.parseInt(ip[0]);
			int m=Integer.parseInt(ip[1]);
			int[] arr=new int[n];
			float[] pro=new float[m];
			int[] idx=new int[m];
			ip=sc.readLine().split(" ");
			for(int i=0;i<n;i++){
				arr[i]=Integer.parseInt(ip[i]);
			}
			for(int i=0;i<m;i++){
				ip=sc.readLine().split(" ");
				pro[i]=Float.parseFloat(ip[1]);
				idx[i]=Integer.parseInt(ip[0]);
			}
			int id=0;
			for(int i=n;i>0;i--){
				if(arr[i-1]!=i){
					id=i;
					break;
				}
			}

			if(id==0) {
				out.println(1.0);
				continue;
			}
			
			double ans=1.0;
			boolean poss=false;
			for(int i=0;i<m;i++){
				if(idx[i]<id) continue;
				ans*=(1.0-pro[i]);
				poss=true;
			}
			if(!poss){
				out.println(0.0);
				continue;
			}
			if(ans!=1.0) out.println((float)(1.0-ans));
			else out.println(1.0);

		}
		out.flush();
		out.close();
		sc.close();
	}

	static boolean isSorted(int arr[],int i,int j){
		boolean sorted=true;
		while(i<j){
			if(arr[i]>arr[i+1]){
				sorted=false;
				break;
			}
			i++;
		}
		return sorted;
	}


}