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).

<image>

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.PrintWriter;
import java.util.StringTokenizer;

public class TaskC {

    public static void main(String[] arg) {

        final FastScanner in = new FastScanner(System.in);
        final PrintWriter out = new PrintWriter(System.out);
        final int T = in.nextInt();
        for(int t=0;t<T;t++){
            final int n = in.nextInt();
            final int m = in.nextInt();
            final int [] a =in.readIntArr(n);
            final int [] el = new int[m];
            final double [] prob = new double[m];
            for(int i=0;i<m;i++){
                el[i] = in.nextInt();
                prob[i] = in.nextDouble();
            }
            Solution sol = new Solution(n, m, a, el, prob);
            double res = sol.solution();
            out.println(res);
        }
        out.close();
        in.close();
    }

    private static class Solution {
        final int n;
        final int m;
        final int []a;
        final int []el;
        final double [] prob;

        public Solution(int n, int m, int[] a, int[] el, double[] prob) {
            this.n = n;
            this.m = m;
            this.a = a;
            this.el = el;
            this.prob = prob;
        }

        private double solution(){
            int max = -1;
            double p = 1.0;
            for(int i=n-1;i>=0;i--){
                if(a[i]!=i+1){
                    max=i;
                    break;
                }
            }
            if(max==-1) return 1.0;
            for(int i=0;i<m;i++){
                if(max<=el[i]-1){
                    p*=(1.0-prob[i]);
                }
            }
            return 1.0-p;
        }

        private double solutionBF(){
            double [] dp = new double[n];
            for(int i=0;i<n;i++){
                if(a[i]==i+1){
                    dp[i]=0;
                }else{
                    dp[i]=1;
                }
            }
            for(int i=0;i<m;i++){
                for(int j=0;j<el[i];j++){
                    dp[j] = dp[j]*(1.0-prob[i]);
                }
            }
            double result = 1.0;
            for(int i=0; i<n; i++){
                result=Math.min(result,1.0-dp[i]);
            }
            return result;
        }
    }

    private static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

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

        int nextInt() {
            return Integer.parseInt(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }

        int[] readIntArr(int n) {
            int[] result = new int[n];
            for (int i = 0; i < n; i++) {
                result[i] = Integer.parseInt(next());
            }
            return result;
        }

        void close() {
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

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

    }


}