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.*;

/**
    @author KhanhNguyenn
 */
 
public class H_bis{
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve();
        out.close();
    }
    // main solver
    static class Task{
        public void solve() {
            Scanner sc= new Scanner(System.in);
            int t= sc.nextInt();
            while(t-->0){
                int n= sc.nextInt(), m= sc.nextInt();
                int[] a= new int[n];
                Pair[] p= new Pair[m];
                for(int i=0;i<n;i++) a[i]=sc.nextInt();
                for(int i=0;i<m;i++){
                    p[i]= new Pair(sc.nextInt(),sc.nextDouble());
                }
                double ans=0;
                double loss=1.000000;
                int j=-1;
                for(int i=n;i>=1;i--){
                    if (a[i-1]!=i){
                        j=i; break;
                    }
                }
                if (j == -1){
                    System.out.println("1.000000");
                    continue;
                }
                for (int i=0;i<m;i++){
                    if(p[i].x < j) continue;
                    ans = (ans + p[i].y * loss);
                    loss *= (1 - p[i].y);
                }
                System.out.printf("%.6f\n",ans);
            }
        }
    }
    static class Pair {
        public int x;
        public double y;
        public Pair(int x, double y){
            this.x = x;
            this.y = y;
        }
 
        // @Override
        // public int compareTo(Pair o) {
        //     if (this.x > o.x){
        //         return 1;
        //     }
        //     else if (this.x < o.x){
        //         return -1;
        //     }
        //     else{
        //         return Integer.compare(this.y, o.y);
        //     }
        // }
    }
    // fast input reader class;
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;
 
        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }
 
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }
 
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public double nextDouble(){
            return Double.parseDouble(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
    }
}