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.*;

public class rand {
    static PrintWriter out;
    static class FIO {
        BufferedReader in;
        StringTokenizer st;
        FIO() throws IOException {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }
        FIO(String name) throws IOException {
            in = new BufferedReader(new FileReader(name + ".in"));
            out = new PrintWriter(new FileWriter(name + ".out"));
        }
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int ni() {
            return Integer.parseInt(next());
        }
        long nl() {
            return Long.parseLong(next());
        }
        double nd() {
            return Double.parseDouble(next());
        }
    }
    static FIO in;
    public static void main (String[] args) throws IOException {
        in = new FIO();
        int t = in.ni();
        while (t-- > 0)
            new solver();
        out.close();
    }
    static class solver {
        solver() {
            int N, M;
            N = in.ni();
            M = in.ni();
            int[] ar = new int[N];
            for (int i = 0; i < N; ++i)
                ar[i] = in.ni();
            int[] srt = new int[N];
            for (int i = 0; i < N; ++i)
                srt[i] = ar[i];
            Arrays.sort(srt);
            int ptr = 0;
            for (int i = N-1; i >= 0; --i)
                if (ar[i] != srt[i]) {
                    ptr = i+1;
                    break;
                }
            double res = 1;
            for (int i = 0; i < M; ++i) {
                int a = in.ni();
                double d = in.nd();
                if (a >= ptr) {
                    res *= (1-d);
                }
            }
            if (ptr == 0) {
                out.println(1);
                return;
            }
            out.println(1-res);
        }
    }
}