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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;

/**
 * @author b
 *
 */
public class C {

	private static Scanner scanner;
	private static PrintWriter writer;

	public static void main(String[] args) {
		scanner = new Scanner();
		writer = new PrintWriter(System.out);
		int tests = scanner.nextInt();
		for (int i = 0; i < tests; i++) {
			solve();
		}
		writer.close();
	}

	private static void solve() {

		int n = scanner.nextInt(),
				ops = scanner.nextInt();

		int[] nums = scanner.nextIntArr(n);

		List<Integer> input = new ArrayList<>(n);
		List<Op> operations = new ArrayList<>();
		for (int num : nums) input.add(num);
		Collections.sort(input);

		for (int i=0; i < ops; i++) {
			operations.add(new Op(scanner.nextInt(), scanner.nextDouble()));
		}

		if (n == 1) {
			printResult(1.0);
			return;
		}

		int wrongPrefix = n;
		for (int i=n-1; i >= 0; i--) {
			if (input.get(i) == nums[i]) wrongPrefix--;
			else break;
		}

		if (wrongPrefix == 0) {
			printResult(1.0);
			return;
		}


		double result = 1.0;
		for (Op operation : operations) {
			if (operation.lengthPrefix >= wrongPrefix) {
				result *= (1.0 - operation.prob);
			}
		}
		printResult(1.0 - result);


	}

	private static void printResult(double result) {
		writer.println(String.format("%.6f", Double.valueOf(result)));
	}

	private static class Op {
		int lengthPrefix;
		double prob;

		public Op(int lengthPrefix, double prob) {
			super();
			this.lengthPrefix = lengthPrefix;
			this.prob = prob;
		}
	}

	@SuppressWarnings("unused")
	private static void print(String... args) {

		for (String arg : args) {

			writer.print(arg + " ");
		}
		writer.println();
	}

	// Props to SecondThread
	static class Scanner {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer tokenizer = new StringTokenizer("");

		String next() {
			while (!tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return tokenizer.nextToken();
		}

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

		long[] nextLongArr(int num) {
			long[] vals = new long[num];
			for (int i = 0; i < num; i++)
				vals[i] = nextLong();
			return vals;
		}

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

		int[] nextIntArr(int numberOfElements) {
			int[] values = new int[numberOfElements];
			for (int i = 0; i < numberOfElements; i++)
				values[i] = nextInt();
			return values;
		}

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

}