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.*;
import java.text.DecimalFormat;

public class _689C
{
    static class Pair implements Comparable<Pair>
    {
        int x; double y;
        Pair(int x, double y) {
            this.x = x;
            this.y = y;
        }
        @Override
        public int compareTo(Pair p2) {
            return this.x - p2.x;
        }
    }
    public static void main(String[] args)
    {
        sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));

        int t = sc.nextInt();
        DecimalFormat df = new DecimalFormat("#.########");
        for (int x = 0; x < t; ++x) {
            int n = sc.nextInt(); int m = sc.nextInt();
            int[] origArr = new int[n]; int[] sortedArr = new int[n];
            for (int i = 0; i < n; ++i) {
                sortedArr[i] = origArr[i] = sc.nextInt();
            }
            Pair[] probs = new Pair[m];
            for (int i = 0; i < m; ++i) {
                probs[i] = new Pair(sc.nextInt()-1, sc.nextDouble());
            }
            Arrays.sort(sortedArr);
            int lastUnsorted = -1;
            for (int i = 0; i < n; ++i)
                if (origArr[i] != sortedArr[i])
                    lastUnsorted = i;
            if (lastUnsorted == -1) {
                out.println(df.format(1));
                continue;
            }
            Arrays.sort(probs);
            double a = 0, probLeft = 1;
            for (int i = 0; i < probs.length && probLeft != 0; ++i) {
                if (probs[i].x >= lastUnsorted) {
                    a += probs[i].y * probLeft;
                    probLeft -= probs[i].y * probLeft;
                }
            }
            out.println(df.format(a));
        }
        out.close();
    }
 
    public static void scan(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            arr[i] = sc.nextInt();
        }
    }
 
    public static MyScanner sc;
    public static PrintWriter out;
 
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public MyScanner() {
            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;
        }
    }
}