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.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;


public class Main{



    static long MOD = 1_000_000_007L;
    //static long MOD = 998_244_353L;
    //static long MOD = 1_000_000_033L;
    static long [] fac;
    static long [] pow;
    static long [] inv;
    static int[][] dir = new int[][]{{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    static long lMax = 0x3f3f3f3f3f3f3f3fL;
    static int iMax = 0x3f3f3f3f;
    static HashMap <Long, Long> memo = new HashMap();
    static MyScanner sc = new MyScanner();
    //static ArrayList <Integer> primes;

    static long[] pow2;
    public static void main(String[] args) {
        out = new PrintWriter(new BufferedOutputStream(System.out));
        // Start writing your solution here. -------------------------------------
        int t = sc.nextInt();
        //int t = 1;

        int nn = 10000;
        /*fac = new long[nn + 1];
        fac[1] = 1;
        for(int i = 2; i <= nn; i++)
            fac[i] = fac[i - 1] * 1L * i % MOD;*/

        /*pow = new long[nn + 1];
        pow[0] = 1L;
        for(int i = 1; i <= nn; i++)
            pow[i] = pow[i - 1] * 2L % MOD;
        pow[1] = 3L;*/

        /*pow2 = new long[nn + 1];
        pow2[0] = 1L;
        for(int i = 1; i <= nn; i++)
            pow2[i] = pow2[i - 1] * 2L % (MOD - 1);*/

        /*inv = new long[nn + 1];
        inv[1] = 1;
        for (int i = 2; i <= nn; ++i)
            inv[i] = (MOD - MOD / i) * inv[(int)(MOD % i)] % MOD;*/


        //primes = sieveOfEratosthenes(100001);

        // 14, [1, 2, 2, 4, 5, 5, 4, 8, 8, 9, 9, 12, 12], [2, 3, 4, 5, 6, 7, 8, 9, 12, 10, 11, 13, 14]
        while(t-- > 0){
            //Solution sol = new Solution();
            //out.println(sol.minDist(5,5,1,5,new int[][]{{1,2,4,2},{1,5,10,3},{3,5,6,3},{2,4,4,3},{4,5,2,1}}));
            Object res = solve();
            out.println(res);
        }
        out.close();
    }
    static boolean [] visited;
    static ArrayList<Integer>[] neigh;
    static int loopRoot;
    static ArrayList <Integer> cycle;
    static boolean loopEnd;
    static long cnt;

    static int total = 0;
    static long inv2 = (MOD + 1) / 2;



    static double solve() {
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] a = new int[n];
        for(int i = 0; i < n; i++)
            a[i] = sc.nextInt();
        int last = n;
        double res = 1.0;
        for(; last > 0; last --)
            if(a[last - 1] != last)
                break;
        if(last == 0) res = 0.0;
        for(int i = 0; i < m; i++){
            int ind = sc.nextInt();
            double prob = sc.nextDouble();
            if(ind >= last)
                res *= (1.0 - prob);
        }

        return 1.0 - res;
    }
    static boolean f(int n){
        int res = 0;
        boolean have[] = new boolean[10];
        while(n != 0) {
            int tmp = n % 10;
            have[tmp] = true;
            res += tmp;
            n /= 10;
        }
        if(res == 12 && res % 2 == 0 && have[res / 2])
            return true;
        return false;
    }


    public static double dist(double a, double b){
        return Math.sqrt(a * a + b * b);
    }

    public static long inv(long a){
        return quickPOW(a, MOD - 2);
    }

    public class Interval {
        int start;
        int end;
        public Interval(int start, int end) {
            this.start = start;
            this.end = end;
        }
    }

    public static ArrayList<Integer> sieveOfEratosthenes(int n) {
        boolean prime[] = new boolean[n + 1];
        Arrays.fill(prime, true);
        for (int p = 2; p * p <= n; p++) {
            if (prime[p]) {
                for (int i = p * 2; i <= n; i += p) {
                    prime[i] = false;
                }
            }
        }
        ArrayList<Integer> primeNumbers = new ArrayList<>();
        for (int i = 2; i <= n; i++) {
            if (prime[i]) {
                primeNumbers.add(i);
            }
        }
        return primeNumbers;
    }


    public static int lowerBound(int[] a, int v){ return lowerBound(a, 0, a.length, v); }
    public static int lowerBound(int[] a, int l, int r, int v)
    {
        if(l > r || l < 0 || r > a.length)throw new IllegalArgumentException();
        int low = l-1, high = r;
        while(high-low > 1){
            int h = high+low>>>1;
            if(a[h] >= v){
                high = h;
            }else{
                low = h;
            }
        }
        return high;
    }
    public static long C(int n, int m)
    {
        if(m == 0 || m == n) return 1l;
        if(m > n || m < 0) return 0l;
        long res = fac[n] * quickPOW((fac[m] * fac[n - m]) % MOD, MOD - 2) % MOD;

        return res;
    }
    public static long quickPOW(long n, long m)
    {
        long ans = 1l;
        while(m > 0)
        {
            if(m % 2 == 1)
                ans = (ans * n) % MOD;
            n = (n * n) % MOD;
            m >>= 1;
        }
        return ans;
    }

    public static int gcd(int a, int b)
    {
        if(a % b == 0) return b;
        return gcd(b, a % b);
    }
    public static long gcd(long a, long b)
    {
        if(a % b == 0) return b;
        return gcd(b, a % b);
    }
    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;
    //-----------MyScanner class for faster input----------
    public 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;
        }
    }
    //--------------------------------------------------------
}