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.*;
import java.lang.reflect.Array;
import java.util.*;
import java.util.stream.Collectors;

public class Main {

    static FastScanner sc;
    static PrintWriter pw;
    static final int INF = 1000000000;
    static final long MOD = 998244353;
    static class A {
        int val; double p;
        int pos;
        A(int x, double y, int pos) {
            this.val = x;
            this.p = y;
            this.pos = pos;
        }

        @Override
        public String toString() {
            return "[ " + val + " " + p + " " + pos + " ]";
        }
    }

    public static void main(String[] args) throws IOException {
        sc = new FastScanner(System.in);
        pw = new PrintWriter(System.out);
        int tt = sc.nextInt();
        for(int ff = 0; ff < tt; ff++) {
            int n = sc.nextInt(), m = sc.nextInt();
            int[] a = new int[n];
            for(int i = 0; i < n; ++i)
                a[i] = sc.nextInt();

            List<A> f = new ArrayList<>();
            for(int i = 0; i < m; ++i) {
                int r = sc.nextInt();
                r--;
                double p = sc.nextDouble();
                f.add(new A(r,p,i));
            }

            Collections.sort(f, (A a1, A a2) -> {
                if(a1.val != a2.val) return a1.val - a2.val;
                else return a1.pos - a2.pos;
            });

            double[] sum = new double[n];
            int x = m - 1;
            for(int i = n - 1; i >= 0; --i) {
                if(i + 1 < n)
                    sum[i] = sum[i + 1];

                while(x >= 0 && f.get(x).val >= i) {
                    sum[i] = sum[i] + ((1.0d - sum[i]) * f.get(x).p);
                    x--;
                }
            }

            double ans = 1.0d;
            for(int i = 0; i < n; ++i)
                if(a[i] - 1 != i) {
                    ans = Math.min(ans, sum[i]);
                }

            pw.println(ans);





        }
        pw.close();

    }
}

class FastScanner {
    static StringTokenizer st = new StringTokenizer("");
    static BufferedReader br;

    public FastScanner(InputStream inputStream) {
        br = new BufferedReader(new InputStreamReader(inputStream));
    }

    public String next() throws IOException {
        while (!st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

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

    public double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

}