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.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Collections;
import java.util.HashMap;
import java.util.Scanner;
import java.util.Set;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws Exception{
		// TTTTTFFFFFF -> freq removal
		Scanner s = new Scanner(System.in);
		int tc = s.nextInt();
		while(tc-- > 0) {
			int n = s.nextInt() , m = s.nextInt();
			int[] input = new int[n];
			int[] temp = new int[n];
			for(int i = 0;i < n;i++) {
				input[i] = s.nextInt();
				temp[i] = input[i];
			}
			int lastcorrectpos = n - 1;
			double ans = 1.0;
			for(int i = n - 1;i >= 0;i--) {
				if(input[i] == lastcorrectpos + 1) lastcorrectpos--;
				else break;
			}
			if(lastcorrectpos == -1) ans = 0;
			for(int i = 0;i < m;i++) {
				int key = s.nextInt() - 1;
				double val = s.nextDouble();
				
				if(key >= lastcorrectpos) ans = ans * (1 - val);
			}
			
			System.out.format("%.6f" , 1 - ans);
			System.out.println();
//			Arrays.sort(input);
//			int index = -1;
//			for(int i = n - 1;i >= 0;i--) {
//				if(input[i] == temp[i]) index = i;
//				else {
//					index = Math.max(0, i);
//					break;
//				}
//			}
//			int si = -1;
//			boolean occur = false;
//			for(int i = 0;i < n;i++) {
//				if(input[i] != temp[i]) {
//					si = i;
//					break;
//				}
//				occur = true;
//			}
//			if(occur && si == -1) {
//				System.out.println("1.000000");
//				continue;
//			}
//			HashMap<Integer , Float> map = new HashMap<>();
//			for(int i = 0;i < m;i++) {
//				int key = s.nextInt();
//				float val = s.nextFloat();
//				map.put(key , val);
//			}
//			int max = temp[0];
//			int[] store = new int[n];
//			for(int i = 0;i < n;i++) {
//				if(max == input[i]) {
//					store[i] = 1;
//					if(i < n - 1) max = temp[i + 1];
//				}
//				else max = Math.max(max, temp[i]);
//			}
//			double mul = 1 , ans = 0;
//			Set<Integer> keys = map.keySet();
//			for(int i : keys) {
//				if(store[i - 1] == 1 && index <= i - 1) {
//					double p = mul * map.get(i);
//					ans += p;
//					mul *= (1 - map.get(i));
//				}
//			}
//			System.out.format("%.6f" , ans);
//			System.out.println();
		}
	}
	
	



	public static int fun(int a , int b) {
		if(a == 0) return 0;
		if(b > a) return 1;

		int val = 1 + fun(a/b , b);
		int val1 = 1 + fun(a , b + 1);

		return Math.min(val , val1);
	}


	public static boolean findDivisor(long n) {
		for(int i = 2;i <= Math.sqrt(n);i++) {
			if(n % i == 0) {
				if(i % 2 == 1) return true;
				else if((n/i) % 2 == 1) return true;
			}
		}
		return false;
	}
	public static void mergeSort(int[] input) {
		mergeSort(input , 0 , input.length - 1);
	}

	public static void mergeSort(int[] input , int si , int ei) {
		if(si >= ei) {
			return;
		}

		int mid = (si + ei) / 2;
		mergeSort(input , si , mid);
		mergeSort(input , mid + 1 , ei);
		merge(input , si , ei);
	}

	public static void merge(int[] input , int si , int ei) {
		int[] temp = new int[ei - si + 1];
		int mid = (si + ei) / 2;
		int i = si;
		int j = mid + 1;
		int k = 0;
		while(i <= mid && j <= ei) {
			if(input[i] < input[j]) {
				temp[k] = input[i];
				i++;
				k++;
			}
			else {
				temp[k] = input[j];
				j++;
				k++;
			}
		}
		while(i <= mid) {
			temp[k] = input[i];
			i++;
			k++;
		}
		while(j <= ei) {
			temp[k] = input[j];
			j++;
			k++;
		}

		for(i = 0;i < temp.length;i++) {
			input[i + si] = temp[i];
		}
	}

}

class pair {
	int h;
	int w;
}
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; 
	} 
}