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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Practice {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		while (t-- > 0) {
			
			String s2=br.readLine();
			int n = Integer.parseInt(s2.split(" ")[0]);
			int k = Integer.parseInt(s2.split(" ")[1]);
			long []arr=new long[n];
		    String str=(br.readLine());
		    String[] s1=str.split(" ");
		    for(int i=0;i<n;i++){
		        arr[i]=Long.parseLong(s1[i]);
		    }
		    Node[]arr2=new Node[k];
		    for(int i=0;i<k;i++) {
		    	String h=br.readLine();
		    	arr2[i]=new Node(Long.valueOf(h.split(" ")[0]), Double.valueOf(h.split(" ")[1]));
		    }
		    Arrays.sort(arr2,new Comparator<Node>() {
		    	public int compare(Node n1,Node n2) {
		    		return (int)(n1.r-n2.r);
		    	}
			});
			System.out.println(getAns(n, arr,arr2,k));
		}
	}

	private static double getAns(long n, long[] arr1,Node[]arr2,long k) {
		// TODO Auto-generated method stub
		long i=arr1.length-1;
		for(i=arr1.length-1;i>=0;i--) {
			if(arr1[(int)i]!=i+1) {
				break;
			}	
		}
		if(i<=0) {
			return (double)1;
		}//System.out.println(i);
		int j=0;
		while(j<arr2.length&&arr2[j].r<i+1) {
			j++;
		}
		if(j==arr2.length) {
			return (double)0;
		}double ans=0;
		double pre=1;
		for(i=j;i<arr2.length;i++) {
			ans=ans+pre*arr2[(int)i].p;
			pre=pre*(1-arr2[(int)i].p);
		}return ans;
	}
}
class Node{
	long r;double p;
	public Node(long r, double p) {
		this.r=r;
		this.p=p;
	}
}