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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class Main {
    static final MyScanner in = new MyScanner();
    static final MyWriter myOut = new MyWriter();
    static final PrintWriter out = myOut.out;

    public static void main(String[] args) {
        int n = in.nextInt();
        while (n-- > 0) {
            int m = in.nextInt();
            int k = in.nextInt();
            int[] arr = new int[m];
            in.nextIntegerArray(arr);
            int number = m + 1;
            for (int i = m - 1; i >= 0; i --) {
                if (arr[i] == i + 1) {
                    number = i + 1;
                }else {
                    break;
                }
            }
            if (number == 1) {
                out.printf("%.6f\n",1.0);
                while (k-- > 0) {
                    in.nextInt();
                    in.nextDouble();
                }
            }else {
                double pre = 1.0;
                double res = 0.0;
                while (k-- > 0) {
                    int x = in.nextInt();
                    double dd = in.nextDouble();
                    //if number = 3,  x = 2 can work
                    if (x >= number - 1) {
                        res += dd * pre;
                        pre *= (1 - dd);
                    }
                }
                out.printf("%.6f\n",res);
            }

        }
        out.flush();
        out.close();

    }

    static class MyOptions {
        static int[] findMax(int[] arr) {
            int max = Integer.MIN_VALUE;
            int index = -1;
            for (int i = 0; i < arr.length; i++) {
                if (max < arr[i]) {
                    index = i;
                    max = arr[i];
                }
            }
            return new int[]{max, index};
        }
        static int[] findMin(int[] arr) {
            int min = Integer.MAX_VALUE;
            int index = -1;
            for (int i = 0; i < arr.length; i++) {
                if (min > arr[i]) {
                    index = i;
                    min = arr[i];
                }
            }
            return new int[]{min, index};
        }
    }

    private static class MyWriter {

        private PrintWriter out;

        public MyWriter() {
            out = new PrintWriter(System.out);
        }

        void printlnArray(int[] arr, int start, int end) {
            for (int i = start; i < end; i++) {
                out.println(arr[i]);
            }
        }

        void printArrayJoinSpace(int[] arr, int start, int end) {
            for (int i = start; i < end; i++) {
                out.println(arr[i] + " ");
            }
        }

        void printlnArray(int[] arr) {
            for (int x : arr) {
                System.out.println(x);
            }
        }

        void printArrayJoinSpace(int[] arr) {
            for (int i : arr) {
                System.out.print(i + " ");
            }
        }

    }

    private static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


        public static int leastInterval(char[] tasks, int n) {
            int i = 0;
            do {
                System.out.println(1);
            } while (i++ < 10);

            String[] split = "0.1".split("\\.");
            System.out.println(Arrays.toString(split));
            return 0;
        }

        public void nextIntegerArray(int[] arr, int start, int end) {
            for (int i = start; i < end; i++) {
                arr[i] = nextInt();
            }
        }

        public void nextIntegerArray(int[] arr) {
            for (int i = 0; i < arr.length; i++) {
                arr[i] = nextInt();
            }
        }
    }


}