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.InputStreamReader;
import java.util.Arrays;

public class C {
    public static void main(String[] args) {
        for (int i = 0; i < 100000; i++) {
            //System.err.println("3 0.000001");
        }
        try {
            BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
            int t = Integer.parseInt(in.readLine());
            for (int i = 0; i < t; i++) {
                String[] s = in.readLine().split(" ");
                int n = Integer.parseInt(s[0]);
                int m = Integer.parseInt(s[1]);
                int[] v = new int[n];
                s = in.readLine().split(" ");
                for (int j = 0; j < n; j++) {
                    v[j] = Integer.parseInt(s[j]);
                }
                int k = 0;
                for (int j = n - 1; j >= 0; j--) {
                    if (v[j] != j + 1) {
                        k = j + 1;
                        break;
                    }
                }
                if (k == 0) {
                    for (int j = 0; j < m; j++) {
                        in.readLine();
                    }
                    System.out.println(1);
                }
                else {
                    double ret = 1;
                    for (int j = 0; j < m; j++) {
                        s = in.readLine().split(" ");
                        int c = Integer.parseInt(s[0]);
                        double p = Double.parseDouble(s[1]);
                        if (c >= k) {
                            ret *= 1-p;
                        }
                    }
                    System.out.println(1 - ret);
                }
            }
            in.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    static int[] c(byte[][] v) {
        int n = v.length;
        int[][] x0 = new int[10][n];
        int[][] x1 = new int[10][n];
        int[][] y0 = new int[10][n];
        int[][] y1 = new int[10][n];
        for (int i = 0; i < 10; i++) {
            Arrays.fill(x0[i], -1);
            Arrays.fill(x1[i], -1);
            Arrays.fill(y0[i], -1);
            Arrays.fill(y1[i], -1);
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int a = v[i][j];
                int[] x0a = x0[a];
                int[] x1a = x1[a];
                if (x0a[i] == -1) x0a[i] = j;
                x1a[i] = j;
            }
        }
        for (int j = 0; j < n; j++) {
            for (int i = 0; i < n; i++) {
                int a = v[i][j];
                int[] y0a = y0[a];
                int[] y1a = y1[a];
                if (y0a[j] == -1) y0a[j] = i;
                y1a[j] = i;
            }
        }
        int[] ret = new int[10];
        for (int i = 0; i < 10; i++) {
            int[] x0a = x0[i];
            int[] x1a = x1[i];
            int[] y0a = y0[i];
            int[] y1a = y1[i];
            int max = 0;
            int xMin = -1;
            for (int j = 0; j < n; j++) {
                if (x0a[j] >= 0) {
                    xMin = j;
                    break;
                }
            }
            int xMax = -1;
            for (int j = n - 1; j >= 0; j--) {
                if (x0a[j] >= 0) {
                    xMax = j;
                    break;
                }
            }
            if (xMin >= 0) {
                for (int j = 0; j < n; j++) {
                    int a = x0a[j];
                    int b = x1a[j];
                    if (a >= 0) {
                        max = Math.max(max, (b - a) * j);
                        max = Math.max(max, (b - a) * (n - j - 1));
                        max = Math.max(max, b * (j - xMin));
                        max = Math.max(max, b * (xMax - j));
                        max = Math.max(max, (n - 1 - a) * (j - xMin));
                        max = Math.max(max, (n - 1 - a) * (xMax - j));
                    }
                }
            }
            int yMin = -1;
            for (int j = 0; j < n; j++) {
                if (y0a[j] >= 0) {
                    yMin = j;
                    break;
                }
            }
            int yMax = -1;
            for (int j = n - 1; j >= 0; j--) {
                if (y0a[j] >= 0) {
                    yMax = j;
                    break;
                }
            }
            if (yMin >= 0) {
                for (int j = 0; j < n; j++) {
                    int a = y0a[j];
                    int b = y1a[j];
                    if (a >= 0) {
                        max = Math.max(max, (b - a) * j);
                        max = Math.max(max, (b - a) * (n - j - 1));
                        max = Math.max(max, b * (j - yMin));
                        max = Math.max(max, b * (yMax - j));
                        max = Math.max(max, (n - 1 - a) * (j - yMin));
                        max = Math.max(max, (n - 1 - a) * (yMax - j));
                    }
                }
            }
            ret[i] = max;
        }
        return ret;
    }
}