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:
//189301019.akshay
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.Random;
import java.util.Arrays;
import java.util.StringTokenizer; 
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Collections;

public class C 
{ 
	static double query[][];
	static double dp[];
	static int n,m,last=0;
    public static void main(String[] args) 
	{ 
		FastReader sc=new FastReader(); 
		StringBuffer ans=new StringBuffer();
		int test=sc.nextInt();
		while(test-->0)
		{
			last=0;
			n=sc.nextInt();
			m=sc.nextInt();
			query=new double[m][2];
			dp=new double[m+5];
			Arrays.fill(dp,-1);
			int arr[]=new int[n];
			int brr[]=new int[n];
			for(int i=0;i<n;i++) {
				arr[i]=sc.nextInt();
				brr[i]=arr[i];
			}
			ruffleSort(brr);
			for(int i=0;i<n;i++) {
				if(arr[i] != brr[i]) last=i+1;
			}
		
			
			for(int i=0;i<m;i++){
				int pos=sc.nextInt();
				double probab=sc.nextDouble();
				query[i][0]=pos;
				query[i][1]=probab;
			}
			if(last == 0) {
				String res1=String.format("%f", 1.0); 
				ans.append(res1+"\n");
				continue;
			}
			double res=solve(0);
			String res1=String.format("%f", res); 
			ans.append(res1+"\n");
		}
		
	    System.out.print(ans);
	} 
    
    static double solve(int qno) {
    	if(qno == m) {
    		return 0.0;
    	}
    	
    	int pos=(int)query[qno][0];
    	double s=query[qno][1];
    	double f=1.0-s;
    	
    	if(dp[qno] != -1) {
    		return dp[qno];
    	}
    	
    	double result=0.0;
    	if(pos>=last) {
    		result=s;
    		result+=(f*solve(qno+1));
    		return dp[qno]=result;
    	}
    	else {
    		result+=solve(qno+1);
    		return dp[qno]=result;
    	}    	
    }
    
    static final Random random=new Random();
    
	static void ruffleSort(int[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n), temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	
	static void ruffleSort(long[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n);
			long temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	
	static class FastReader 
	{ 
		BufferedReader br; 
		StringTokenizer st; 

		public FastReader() 
		{ 
			br = new BufferedReader(new
					InputStreamReader(System.in)); 
		} 

		String next() 
		{ 
			while (st == null || !st.hasMoreElements()) 
			{ 
				try
				{ 
					st = new StringTokenizer(br.readLine()); 
				} 
				catch (IOException e) 
				{ 
					e.printStackTrace(); 
				} 
			} 
			return st.nextToken(); 
		} 

		int nextInt() 
		{ 
			return Integer.parseInt(next()); 
		} 

		long nextLong() 
		{ 
			return Long.parseLong(next()); 
		} 

		double nextDouble() 
		{ 
			return Double.parseDouble(next()); 
		} 

		String nextLine() 
		{ 
			String str = ""; 
			try
			{ 
				str = br.readLine(); 
			} 
			catch (IOException e) 
			{ 
				e.printStackTrace(); 
			} 
			return str; 
		} 
	} 


}