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.util.*;		
import java.util.function.Function;
import java.util.stream.Collectors;
import java.io.*;
import java.math.*;
public class Main7{
	static public void main(String args[])throws IOException{
		int tt=i();
		StringBuilder sb=new StringBuilder();
		for(int ttt=1;ttt<=tt;ttt++){
			int n=i();
			int m=i();
			int[] a=new int[n];
			for(int i=0;i<n;i++){
				a[i]=i();
			}
			int[] r=new int[m];
			double[] val =new double[m];
			for(int i=0;i<m;i++){
				r[i]=i();
				String s=s();
				val[i]=Double.parseDouble(s);
			}
			int count=0;
			for(int i=n-1;i>=0;i--){
				if(a[i]==i+1){
					count++;
				}else{
					break;
				}
			}
			int index=n-count;
			ArrayList<Double> ar=new ArrayList<>();
			for(int i=0;i<m;i++){
				if(r[i]>=index){
					ar.add(val[i]);
				}
			}
			if(index==0){
				sb.append("1.00000\n");
			}else{
	//		pln(index+" "+ar);
			double ans=0.0;
			double tmp=1;
			for(int i=0;i<ar.size();i++){
				ans=ans+tmp*ar.get(i);
				tmp=tmp*(1-ar.get(i));
			}
			sb.append(ans+"\n");}
		}
		System.out.print(sb.toString());
	}
	
	
	static InputReader in=new InputReader(System.in);
	static OutputWriter out=new OutputWriter(System.out);
	static ArrayList<ArrayList<Integer>> graph;
	static int mod=998244353;
	static class Pair{
		long x;
		int y;
		Pair(long x,int y){
			this.x=x;
			this.y=y;
		}
		/*@Override
		public int hashCode() 
		{ 
			final int temp = 14; 
			int ans = 1; 
			ans =x*31+y*13; 
			return (int)ans;
		}
       @Override
        public boolean equals(Object o) 
        { 
            if (this == o) { 
                return true; 
            } 
            if (o == null) { 
                return false; 
            } 
            if (this.getClass() != o.getClass()) { 
                return false; 
            } 
            Pair other = (Pair)o; 
            if (this.x != other.x || this.y!=other.y) { 
                return false; 
            } 
            return true; 
        }*/
	}
	public static long[] sort(long[] a){
		int n=a.length;
		ArrayList<Long> ar=new ArrayList<>();
		for(int i=0;i<a.length;i++){
			ar.add(a[i]);
		}
		Collections.sort(ar);
		for(int i=0;i<n;i++){
			a[i]=ar.get(i);
		}
		return a;
	}
	public static long pow(long a, long b){
        long result=1;
        while(b>0){
            if (b % 2 != 0){
                result=(result*a)%mod;
                b--;
            } 
            a=(a*a)%mod;
            b /= 2;
        }   
        return result;
    }
	public static long gcd(long a, long b){
        if (a == 0){
            return b;
		}
        return gcd(b%a, a);
    }
	public static long lcm(long a, long b){
		return a*(b/gcd(a,b));
	}
	public static long l(){
		String s=in.String();
		return Long.parseLong(s);
	}
	public static void pln(String value){
		System.out.println(value);
	}
	public static int i(){
		return in.Int();
	}
	public static String s(){
		return in.String();
	}
}
class InputReader 
{
	private InputStream stream;
	private byte[] buf = new byte[1024];
	private int curChar;
	private int numChars;
	private SpaceCharFilter filter;
	public InputReader(InputStream stream) {
		this.stream = stream;
	}
	public int read() {
		if (numChars== -1)
			throw new InputMismatchException();
		if (curChar >= numChars) {
			curChar = 0;
			try {
				numChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (numChars <= 0)
				return -1;
		}
		return buf[curChar++];
	}
	public int Int() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
	public String String() {
		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 boolean isSpaceChar(int c) {
		if (filter != null)
			return filter.isSpaceChar(c);
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	} 
	public String next() {
		return String();
	} 
	public interface SpaceCharFilter {
		public boolean isSpaceChar(int ch);
	}
}
class OutputWriter {
	private final PrintWriter writer;
	public OutputWriter(OutputStream outputStream) {
		writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
	}
	public OutputWriter(Writer writer) {
		this.writer = new PrintWriter(writer);
	}
	public void print(Object...objects) {
		for (int i = 0; i < objects.length; i++) {
			if (i != 0)
				writer.print(' ');
			writer.print(objects[i]);
		}
	}
	public void printLine(Object...objects) {
		print(objects);
		writer.println();
	}
	public void close() {
		writer.close();
	}
	public void flush() {
		writer.flush();
	}
 }
class IOUtils {
	public static int[] readIntArray(InputReader in, int size) {
		int[] array = new int[size];
		for (int i = 0; i < size; i++)
			array[i] = in.Int();
		return array;
	}
}