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.lang.*;
import java.io.*; 
import javafx.util.Pair; 
public class Main 
{
//     HashMap<> map=new HashMap<>();
//     TreeMap<> map=new TreeMap<>();
//     map.put(p,map.getOrDefault(p,0)+1);
//     for(Map.Entry<> mx:map.entrySet()){
//           int v=mx.getValue(),k=mx.getKey();
//     }
//     ArrayList<Pair<Character,Integer>> l=new ArrayList<>();
//     ArrayList<> l=new ArrayList<>();
//     HashSet<> has=new HashSet<>();
    PrintWriter out;
    FastReader sc;
    public void sol(){
          int t=ni();
          while(t-->0){
                int n=ni(),m=ni();
                int[] ar=new int[n+1];
                for(int i=1;i<=n;i++){
                      ar[i]=ni();
                }double result=1;
                int j=0;
                for(int i=n;i>=1;i--){
                      if(ar[i]!=i){
                            j=i;break;
                      }
                }if(j==0){
                      result=0;
                }
                for(int i=0;i<m;i++){
                      int a=ni();
                      double b=nd();
                      if(a>=j){
                            result*=(1-b);
                      }
                }pl(1-result);
          }
    }
    public static void main(String[] args) 
	{ 
	      Main g=new Main();
		g.out=new PrintWriter(System.out);
		g.sc=new FastReader();
		g.sol();
		g.out.flush();
	}
	static class FastReader 
	{ 
		BufferedReader br; 
		StringTokenizer st; 
 
		public FastReader() 
		{ 
			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; 
		} 
	} public int ni(){
	      return sc.nextInt();
	}public long nl(){
	      return sc.nextLong();
	}public double nd(){
	      return sc.nextDouble();
	}public String rl(){
	      return sc.nextLine();
	}public void pl(Object s){
	      out.println(s);
	}public void ex(){
	      out.println();
	}
	public void pr(Object s){
	      out.print(s);
	}public String next(){
	      return sc.next();
	}public long abs(long x){
	      return Math.abs(x);
	}
	public int abs(int x){
	      return Math.abs(x);
	}
	public double abs(double x){
	      return Math.abs(x);
	}
      public long pow(long x,long y){
            return (long)Math.pow(x,y);
      }
      public int pow(int x,int y){
            return (int)Math.pow(x,y);
      }
      public double pow(double x,double y){
            return Math.pow(x,y);
      }public long min(long x,long y){
            return (long)Math.min(x,y);
      }
      public int min(int x,int y){
            return (int)Math.min(x,y);
      }
      public double min(double x,double y){
            return Math.min(x,y);
      }public static int gcd(int a, int b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a) {
			l.add(i);
		}
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) {
			a[i] = l.get(i);
		}
	}void sort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a) {
			l.add(i);
		}
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) {
			a[i] = l.get(i);
		}
	}void sort(double[] a) {
		ArrayList<Double> l = new ArrayList<>();
		for (double i : a) {
			l.add(i);
		}
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) {
			a[i] = l.get(i);
		}
	}int  swap(int a,int b){
	      return a;
	}long  swap(long a,long b){
	      return a;
	}double  swap(double a,double b){
	      return a;
	}
	boolean isPowerOfTwo (int x) 
      { 
            return x!=0 && ((x&(x-1)) == 0);     
      }boolean isPowerOfTwo (long x) 
      { 
            return x!=0 && ((x&(x-1)) == 0);     
      }public long max(long x,long y){
            return (long)Math.max(x,y);
      }
      public int max(int x,int y){
            return (int)Math.max(x,y);
      }
      public double max(double x,double y){
            return Math.max(x,y);
      }long sqrt(long x){
            return (long)Math.sqrt(x);
      }int sqrt(int x){
            return (int)Math.sqrt(x);
      }
	 
}