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.util.*;
import java.math.*;

import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.OutputStream;

public class c {
    public static void main(String[] args) {
        Kattio io = new Kattio(System.in, System.out);

        int t = io.getInt();
        while (t-- > 0) {
            int n = io.getInt(); int m = io.getInt();
            int lastWrong = 0;
            for (int i = 1; i <= n; i++) {
                if (io.getInt() != i) lastWrong = i;
            }

            double P = 1;
            for (int i = 0; i < m; i++) {
                int r = io.getInt(); double p = io.getDouble();
                if (r >= lastWrong) P *= (1 - p);
            }
            io.println(lastWrong == 0 ? 1 : 1 - P);
        }
        io.flush();
    }
}


// ------------------------------------------------------------

class Kattio extends PrintWriter {
    public Kattio(InputStream i) {
        super(new BufferedOutputStream(System.out));
        r = new BufferedReader(new InputStreamReader(i));
    }
    public Kattio(InputStream i, OutputStream o) {
        super(new BufferedOutputStream(o));
        r = new BufferedReader(new InputStreamReader(i));
    }

    public boolean hasMoreTokens() {
        return peekToken() != null;
    }

    // hasMoreLineTokens and getLine from
    // https://github.com/johanstenberg92/SHA-256/blob/master/src/org/johanstenberg/sha256/Kattio.java
    public boolean hasMoreLineTokens() throws Throwable {
        if (line == null) {
            line = r.readLine();
        }

        return line != null;
    }

    public String getLine() throws Throwable {
        if (line == null) line = r.readLine();
        if (line != null) {
            String tmp = line;
            line = null;
            return tmp;
        }

        return null;
    }

    public int getInt() {
        return Integer.parseInt(nextToken());
    }

    public double getDouble() {
        return Double.parseDouble(nextToken());
    }

    public long getLong() {
        return Long.parseLong(nextToken());
    }

    public String getWord() {
        return nextToken();
    }



    private BufferedReader r;
    private String line;
    private StringTokenizer st;
    private String token;

    private String peekToken() {
        if (token == null)
            try {
                while (st == null || !st.hasMoreTokens()) {
                    line = r.readLine();
                    if (line == null) return null;
                    st = new StringTokenizer(line);
                }
                token = st.nextToken();
            } catch (IOException e) { }
        return token;
    }

    private String nextToken() {
        String ans = peekToken();
        token = null;
        return ans;
    }
}