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.ArrayList;
import java.util.List;

public class RandomEvents {
    public static void main(String[] args) throws Exception {
        // BufferedReader br = new BufferedReader(new FileReader(new File("input.txt")));
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        handleMultipleTestCases(br);
    }

    private static void handleMultipleTestCases(BufferedReader br) throws Exception {
        int T = readInt(br);
        for (int tc = 0; tc < T; tc++) {
            handleTestCase(br);
        }
    }

    private static void handleTestCase(BufferedReader br) throws Exception {
        int[] inpLine = readArrInt(br);
	int N = inpLine[0];
	int M = inpLine[1];
	int[] arr = readArrInt(br);
	double[][] ran = new double[M][2];
	for(int i=0; i<M; i++) {
		ran[i] = readArrDouble(br);
	}
	getProbability(arr,ran);
    }

    private static int readInt(BufferedReader br) throws Exception {
        return Integer.parseInt(br.readLine());
    }

    private static int[] readArrInt(BufferedReader br) throws Exception {
        String[] line = br.readLine().split(" ");
        int[] arr = new int[line.length];
        for (int i = 0; i < line.length; i++) {
            arr[i] = Integer.parseInt(line[i]);
        }
        return arr;
    }

    private static double[] readArrDouble(BufferedReader br) throws Exception {
        String[] line = br.readLine().split(" ");
        double[] arr = new double[line.length];
        for (int i = 0; i < line.length; i++) {
            arr[i] = Double.parseDouble(line[i]);
        }
        return arr;
    }


    private static List<Integer> readList(BufferedReader br) throws Exception {
        String[] line = br.readLine().split(" ");
        List<Integer> arr = new ArrayList<>();
        for (int i = 0; i < line.length; i++) {
            arr.add(Integer.parseInt(line[i]));
        }
        return arr;
    }

    private static void print(int[][] mat) {
        System.out.println("----------------------------------------");
        for (int i = 0; i < mat.length; i++) {
            System.out.print("{");
            for (int j = 0; j < mat[0].length; j++) {
                System.out.print(mat[i][j] + " ");
            }
            System.out.println("}");
        }
        System.out.println("----------------------------------------");
    }

    private static void print(int[] mat) {
        System.out.print("{");
        for (int i = 0; i < mat.length; i++) {
            System.out.print(mat[i] + " ");
        }
        System.out.println("}");
    }

    private static void getProbability(int[] arr, double[][] ranges) {
    	int n= arr.length;
	int M = arr.length;
	double revP = 1;
	boolean flag = false;

	for(int i=arr.length-1; i>=0; i--) {
		if(arr[i]!=i+1) {
			M = i+1;
			flag = true;
			break;
		}
	}
	if(!flag) {
		System.out.println("1.000000");
	}else {

	for(int i=0; i<ranges.length; i++) {
		if((int)ranges[i][0]>=M) {
			revP = revP * (1-ranges[i][1]);
		}
	}
	System.out.print(String.format("%.6g%n", (1-revP)));
	}
    }
}