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.util.*;
public class check3  {

    public static void main(String[] args) throws IOException{

        Reader sc=new Reader();
        PrintWriter out = new PrintWriter(System.out);
        int t=sc.nextInt();

        while(t-->0)
        {
            int n=sc.nextInt();
            int m=sc.nextInt();
            int arr[]=new int[n];
            int max=-1;

            for(int i=0;i<n;i++)
            {
                arr[i]=sc.nextInt();
                if(arr[i]!=(i+1))
                {
                    max=Math.max(max,arr[i]);
                }
            }
            if(max==-1)
            {
                for(int i=0;i<m;i++) {
                    int ind = sc.nextInt();
                    double d1 = sc.nextDouble();
                }
                out.println(1);
                continue;
            }
            double ans=1;
            ArrayList<Double> res=new ArrayList<>();

            for(int i=0;i<m;i++)
            {
                int ind=sc.nextInt();
                double d1=sc.nextDouble();
                if(ind<max)
                {
                    continue;
                }
                //update ans
                res.add(1-d1);
            }
//            double
            for(int i=0;i<res.size();i++)
            {
                ans*=res.get(i);
            }
            out.println(1-ans);

        }
        out.flush();
    }

    public static long power(long x,long power)
    {
        if(power==0) return 1;

        long ans1 = power(x,power/2);
        ans1=ans1*ans1;

        if(power%2==0) return ans1;
        return x*ans1;
    }

    //    (a*a_inverse) = 1(mod m)
//    finding a_inverse
    public static long modInverse(long a,long m)
    {
//        works only when m is prime
        long g=gcd(a,m);
        if(g==1) return power(a,m-2,m);
        else return -1;
    }

    //    (x^power)mod(m)
    public static long power(long x,long power,long m)
    {
        if(power==0) return 1;

        long p = power(x,power/2,m)%m;
        p = (p*p)%m;

        if(power%2==0) return p;
        return (x*p)%m;
    }
    public static long gcd(long small,long large)
    {
        if(small==0) return large;
        return gcd(large % small,small);
    }

    public static boolean isprime(long no)
    {
        if(no<=1) return false;
        if(no<=3) return true;
        if(no%2==0 || no%3==0) return false;
        // 6k+1 6k+5 can be prime
        for(long i=5;i*i<=no;i+=6)
        {
            if(no%(i)==0 || no%(i+2)==0)
                return false;
        }
        return true;
    }

    //    prime no smaller than or equal to n
    public static boolean[] prime_nos_till_no(int no)
    {
        boolean prime[]=new boolean[no+1];
//        prime[i]== true means prime
//        prime[i]== false means not a prime
//        initialize prime array as true

        Arrays.fill(prime,true);
        prime[0]=false;
        for(int i=2;i*i<=no;i++)
        {
            if(prime[i]==true)
            {
                for(int j=i*i;j<=no;j+=i)
                {
                    prime[j]=false;
                }
            }
        }
        return prime;

    }
    public static void shufflearray(long arr[])
    {


        int n=arr.length;
        Random rand=new Random();
        for(int i=0;i<n;i++)
        {
            long temp=arr[i];
            int randomPos=i+rand.nextInt(n-i);
            arr[i]=arr[randomPos];
            arr[randomPos]=temp;
        }
    }
    //       Arrays.sort(arr, new Comparator<pair>() {
//        //@Override
//        public int compare(pair o1, pair o2) {
//            long l1=o1.a-o2.a;
//            if(l1<0L) return -1;
//            if(l1==0) return 0;
//            return 1;
//        }
//    });
    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String nextLine() throws IOException {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }
        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public String next() throws IOException{
            int c = read();
            while (isSpaceChar(c)) c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }


        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            if (din == null)
                return;
            din.close();
        }

    }

}