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.util.*;
import java.util.function.Function;

public class Main {

    public static void main(String[] args) {

        Parser parser = new Parser();
        int t = parser.parseInt();
        for (int i = 1; i <= t; i++) {
            Problem problem = new Problem();
            problem.solve(i);
        }
    }
}

class Problem {
    private Parser parser = new Parser();

    public void solve() {
    }

    public void solve(int testNum) {
        int n = parser.parseInt();
        int m = parser.parseInt();

        int[] a = new int[n + 1];
        for(int i = 1; i <= n; i++) {
            a[i] = parser.parseInt();
        }

        boolean[] isSorted = new boolean[n + 1];
        isSorted[n] = (a[n] == n);
        for(int i = n - 1; i >= 1; i--) {
            isSorted[i] = isSorted[i + 1] && a[i] == i;
        }

        int firstIndex = 0;
        for(int i = n; i >= 1; i--) {
            if(!isSorted[i]) {
                firstIndex = i;
                break;
            }
        }

        double ans = 0;
        double curr = 1;
        for(int i = 1; i <= m; i++) {
            int index = parser.parseInt();
            double percent = parser.parseDouble();
            if(index >= firstIndex) {
                ans += (curr * percent);
                curr *= (1 - percent);
            }
        }

        if(isSorted[1]) {
            System.out.printf("%.12f\n", 1.0);
            return;
        }

        System.out.printf("%.12f\n", ans);
    }
}

class MathUtil {

    public static List<Integer> getPrimes(int r) {
        if (r < 2) {
            return List.of();
        }

        List<Integer> result = new ArrayList<>();
        result.add(2);
        boolean[] is = new boolean[r + 1];
        for (int i = 4; i <= r; i += 2) {
            is[i] = true;
        }
        for (int i = 3; i <= r; i += 2) {
            if (is[i]) continue;
            result.add(i);
            for (long j = (long) i * i; j <= r; j += i) {
                is[(int) j] = true;
            }
        }
        return result;
    }

    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    public static long pow(long x, long y) {
        return pow(x, y, Function.identity());
    }

    public static long pow(long x, long y, long mod) {
        return pow(x, y, v -> v % mod);
    }

    private static long pow(long x, long y, Function<Long, Long> action) {
        long res = 1;
        long t = x;
        while (y > 0) {
            if ((y & 1) == 1) {
                res *= t;
                res = action.apply(res);
            }
            t *= t;
            t = action.apply(t);
            y /= 2;
        }
        return res;
    }
}

class DisjointSet {

    private final int[] f;

    public DisjointSet(int size) {
        f = new int[size + 1];
        for (int i = 0; i <= size; i++) {
            f[i] = i;
        }
    }

    public int find(int u) {
        if (f[u] == u) return u;
        return f[u] = find(f[u]);
    }

    public boolean merge(int u, int v) {
        u = find(u);
        v = find(v);
        if (u == v) return false;
        if (u > v) {
            int t = u;
            u = v;
            v = t;
        }
        f[u] = v;
        return true;
    }
}

class IntPair {
    private int l;
    private int r;

    public IntPair(int l, int r) {
        this.l = l;
        this.r = r;
    }

    public int getL() {
        return l;
    }

    public void setL(int l) {
        this.l = l;
    }

    public int getR() {
        return r;
    }

    public void setR(int r) {
        this.r = r;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        IntPair pair = (IntPair) o;
        return l == pair.l &&
                r == pair.r;
    }

    @Override
    public int hashCode() {
        return Objects.hash(l, r);
    }
}

class Parser {
    public static final Parser PARSER = new Parser();
    private static final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private StringTokenizer tokenizer;

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

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

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

    public String parseLine() {
        try {
            return reader.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }

    String parseString() {
        fill();
        return tokenizer.nextToken();
    }

    private void fill() {
        while (tokenizer == null || !tokenizer.hasMoreElements()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }
}