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 d2prac implements Runnable 
{
	private boolean console=false;

	public void solve()
	{
		int i; int n=in.ni(),k=in.ni(); Integer a[]=new Integer[n]; double ans=1.0; int max=Integer.MAX_VALUE; boolean sorted=true;
		for(i=0;i<n;i++) 
		{
			a[i]=in.ni()-1; 
			if(a[i]!=i)
			{
				max=i; sorted=false;
			}
		}
		boolean h=false;
		for(i=0;i<k;i++)
		{
			int len=in.ni()-1; double p=in.nd();
			if(len>=max)
			{
				ans*=(1-p); h=true;
			}
		}
		if(sorted) 
		{
			out.println(1); return;
		}
		if(!h)
		{
			out.println(0); return;
		}
		out.println(1-ans);
		
	}
	/*public boolean dfs(int x,int y,int size)
	{
		int fy=y-(size-1) ,ly=y+(size-1); int move=1;
		if(fy<0||ly>=m||x+size-1>=n) return false;
		for(int i=x+1;i<=x+size-1;i++)
		{
			for(int left=0;left<=move;left++)
				if(a[i][y-left]=='.') return false;
			for(int right=0;right<=move;right++)
				if(a[i][y+right]=='.') return false;
			move++;
		}
		return true;
	}*/
	@Override
	public void run() {
		try {  init();  }  
		catch (FileNotFoundException e) {  e.printStackTrace();  }

		int t= in.ni();

		while (t-->0) {
			solve();
			out.flush(); }
	}
	private FastInput in;    private PrintWriter out;
	public static void main(String[] args) throws Exception {	 new d2prac().run();	 }

	private void init() throws FileNotFoundException {
		InputStream inputStream = System.in; 	 OutputStream outputStream = System.out;
		try {  if (!console && System.getProperty("user.name").equals("sachan")) {
			outputStream = new FileOutputStream("/home/sachan/Desktop/output.txt");
			inputStream = new FileInputStream("/home/sachan/Desktop/input.txt");   	}
		}	catch (Exception ignored) {	}
		out = new PrintWriter(outputStream); 	 in = new FastInput(inputStream);
	}
	static class FastInput { InputStream obj;
	public FastInput(InputStream obj) {	this.obj = obj;	}
	byte inbuffer[] = new byte[1024];  int lenbuffer = 0, ptrbuffer = 0;
	int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
	if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
	try { lenbuffer = obj.read(inbuffer);  }
	catch (IOException e) { throw new InputMismatchException(); } }    
	if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }

	String ns() { int b = skip();StringBuilder sb = new StringBuilder();
	while (!(isSpaceChar(b)))	  { sb.appendCodePoint(b);b = readByte(); }return sb.toString();}

	int ni() { int num = 0, b;boolean minus = false;
	while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
	if (b == '-') { minus = true;b = readByte(); }
	while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
		return minus ? -num : num; }b = readByte(); }}

	long nl() { long num = 0;int b;boolean minus = false;
	while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
	if (b == '-') { minus = true;b = readByte(); }
	while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
		return minus ? -num : num; }b = readByte(); } }

	boolean isSpaceChar(int c) { return (!(c >= 33 && c <= 126)); }
	int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
	float nf() {return Float.parseFloat(ns());}
	double nd() {return Double.parseDouble(ns());}
	char nc() {return (char) skip();}
	}
}