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 Main{

	public static Reader sc = new Reader();
	//public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

   public static void main(String[] args) throws IOException{
	
	int t = sc.nextInt();
	while (t-->0){
		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[] r = new int[m];
		double[] p = new double[m];
		for (int i = 0; i < m; i++) {
			r[i] = sc.nextInt()-1;
			p[i] = sc.nextDouble();
		}
		double fp = 1;
		int bi = -1;
		for (int i = n-1; i >= 0; i--) {
			if (a[i] != i+1) {
				bi = i;
				break;
			}
		}
		if (bi == -1) out.println(1);
		else {
			for (int i = 0; i < m; i++) {
				if (r[i] >= bi) {
					fp *= (1-p[i]);
				} 
			}
			out.println(1-fp);
		}
	}
	
	out.close();
   }
   static long ceil(long a, long b) {
		return (a+b-1)/b;
	}
  
   static long powMod(long base, long exp, long mod) {
	   if (base == 0 || base == 1) return base;
	   if (exp == 0) return 1;
	   if (exp == 1) return base % mod;
	   long R = powMod(base, exp/2, mod) % mod;
	   R *= R;
	   R %= mod;
	   if ((exp & 1) == 1) {
		   return base * R % mod;
	   }
	   else return R % mod;
   }
   
   static long pow(long base, long exp) {
	   if (base == 0 || base == 1) return base;
	   if (exp == 0) return 1;
	   if (exp == 1) return base;
	   long R = pow(base, exp/2);
	   if ((exp & 1) == 1) {
		   return R * R * base;
	   }
	   else return R * R;
   }
   

 
   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 readLine() 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, Integer.max(cnt-1, 0)); 
       } 

       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(); 
       } 
   }

	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
}