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 final class Solution{
	static Scanner sc=new Scanner(System.in);
//	static FastReader sc=new FastReader();
    static PrintWriter writer=new PrintWriter(System.out);
    
    static int binarySearch(int[] a, int l, int r, int key) {
    	int in=l, max=r;
    	while(l<=r) {
    		int mid = (l+r)/2;
    		if(a[mid]<=key) {
    			in = Math.min(mid+1, r);
    			l=mid+1;
    		}else r=mid-1;
    	}
    	return Math.min(in+1, max);
    }
    
    static void solve() {
    	int t=sc.nextInt();
    	while(t-->0) {
    		int n=sc.nextInt(), m=sc.nextInt();
    		int[] a=new int[n];
    		for(int i=0;i<n;i++) a[i]=sc.nextInt();
    		int r=n-1;
    		for(int i=n-1;i>=0;i--) {
    			if(a[i]!=i+1) {r=i+1; break;}
    			else r=-1;
    		}
    		double ans=1;
    		for(int i=0;i<m;i++) {
    			int j=sc.nextInt();
    			double p=sc.nextDouble();
    			if(j>=r) {
    				ans*=(1-p);
    			}
    		}
			if(r==-1) ans=0;
			System.out.printf("%.6f\n", 1-ans);
    	}
    }

	public static void main(String[] args){
		int t=0;
		solve();
		while(t-->0) {
			int n=sc.nextInt(), q=sc.nextInt();
			int[] a=new int[n];
			int[] pre=new int[n];
			for(int i=0;i<n;i++) a[i]=sc.nextInt();
			Arrays.sort(a);
			pre[0]=a[0];
			for(int i=1;i<n;i++) pre[i]=pre[i-1]+a[i];
			
			for(int i=0;i<q;i++) {
				int x=sc.nextInt();
				int l=0, r=n-1;
				String ans="No";
				while(l<r) {
					int mid = binarySearch(a, l, r, (a[l]+a[r])/2);
					System.out.println(mid);
					if(mid==l) {
						if(pre[r]==x) ans="Yes";
						else ans="No";
						break;
					}else if((l==0&&pre[mid-1]==x) || (pre[mid-1]-pre[l]==x) || pre[r]-pre[mid-1]==x) {ans="Yes"; break;}
					else if((l==0&&pre[mid-1]>x) || (pre[mid-1]-pre[l]>x)) r=mid-1;
					else if(pre[r]-pre[mid-1]>x) l=mid;
					else {ans="No"; break;}
//					System.out.println(l+" "+mid+" "+r+" "+ans);
				}
				System.out.println(ans);
			}
		}
        writer.flush();
        writer.close();
	}
	
	
  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; 
        } 
    }

}