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.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.*;

/**
 * @author Andrei Chugunov
 */
public class Main {

    private static class Solution {
        private void solve(FastScanner in, PrintWriter out) {
            int t = in.nextInt();
            for (int i = 0; i < t; i++) {
                int n = in.nextInt();
                int m = in.nextInt();
                int[] arr = in.nextIntArray(n);
                int R = n;
                for (int j = n-1; j >= 0; j--) {
                    if (arr[j] == j+1) {
                        R--;
                    } else break;
                }

                double ans = R == 0 ? 0 : 1;
                for (int j = 0; j < m; j++) {
                    int r = in.nextInt();
                    double p = in.nextDouble();
                    if (r < R) {
                        continue;
                    }
                    ans = ans * (1 - p);

                }
                out.println(1 - ans);
            }
        }
    }

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solution solver = new Solution();
        solver.solve(in, out);
        out.flush();
        //out.close();
    }

    private static class ArrayUtils {
        private static final long seed = System.nanoTime();
        private static final Random rand = new Random(seed);

        private static void sort(int[] a) {
            shuffle(a);
            Arrays.sort(a);
        }

        private static void shuffle(int[] a) {
            for (int i = 0; i < a.length; i++) {
                int j = rand.nextInt(i + 1);
                int t = a[i];
                a[i] = a[j];
                a[j] = t;
            }
        }

    }

    private static class FastScanner {
        private BufferedReader br;
        private StringTokenizer st;

        FastScanner(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            st = null;
        }

        private String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

        private int[] nextIntArray(int n) {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }
    }
}