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.Arrays;
import java.util.InputMismatchException;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	static class TaskAdapter implements Runnable {
		@Override
		public void run() {
			long startTime = System.currentTimeMillis();
			InputStream inputStream = System.in;
			OutputStream outputStream = System.out;
			FastReader in = new FastReader(inputStream);
			Output out = new Output(outputStream);
			CRandomEvents solver = new CRandomEvents();
			int testCount = in.nextInt();
			for(int i = 1; i<=testCount; i++)
				solver.solve(i, in, out);
			out.close();
			System.err.println(System.currentTimeMillis()-startTime+"ms");
		}
	}

	public static void main(String[] args) throws Exception {
		Thread thread = new Thread(null, new TaskAdapter(), "", 1<<28);
		thread.start();
		thread.join();
	}

	static class CRandomEvents {
		public CRandomEvents() {
		}

		public void solve(int kase, InputReader in, Output pw) {
			int n = in.nextInt(), m = in.nextInt();
			int[] arr = in.nextInt(n);
			int i = n-1;
			for(; i>=0&&arr[i]==i+1; i--) ;
//			Utilities.Debug.dbg(i);
			double ans = 1;
			for(int j = 0; j<m; j++) {
				int r = in.nextInt();
				double p = in.nextDouble();
				if(r>i) {
					ans *= (1-p);
				}
			}
			if(i==-1) {
				ans = 0;
			}
			pw.printf("%.6f\n", 1-ans);
		}

	}

	interface InputReader {
		int nextInt();

		double nextDouble();

		default int[] nextInt(int n) {
			int[] ret = new int[n];
			for(int i = 0; i<n; i++) {
				ret[i] = nextInt();
			}
			return ret;
		}

	}

	static class FastReader implements InputReader {
		final private int BUFFER_SIZE = 1<<16;
		private final DataInputStream din;
		private final byte[] buffer;
		private int bufferPointer;
		private int bytesRead;

		public FastReader(InputStream is) {
			din = new DataInputStream(is);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public int nextInt() {
			int ret = 0;
			byte c = skipToDigit();
			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() {
			double ret = 0, div = 1;
			byte c = skipToDigit();
			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 boolean isDigit(byte b) {
			return b>='0'&&b<='9';
		}

		private byte skipToDigit() {
			byte ret;
			while(!isDigit(ret = read())&&ret!='-') ;
			return ret;
		}

		private void fillBuffer() {
			try {
				bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			}catch(IOException e) {
				e.printStackTrace();
				throw new InputMismatchException();
			}
			if(bytesRead==-1) {
				buffer[0] = -1;
			}
		}

		private byte read() {
			if(bytesRead==-1) {
				throw new InputMismatchException();
			}else if(bufferPointer==bytesRead) {
				fillBuffer();
			}
			return buffer[bufferPointer++];
		}

	}

	static class Output implements Closeable, Flushable {
		public StringBuilder sb;
		public OutputStream os;
		public int BUFFER_SIZE;
		public String lineSeparator;

		public Output(OutputStream os) {
			this(os, 1<<16);
		}

		public Output(OutputStream os, int bs) {
			BUFFER_SIZE = bs;
			sb = new StringBuilder(BUFFER_SIZE);
			this.os = new BufferedOutputStream(os, 1<<17);
			lineSeparator = System.lineSeparator();
		}

		public void print(String s) {
			sb.append(s);
			if(sb.length()>BUFFER_SIZE >> 1) {
				flushToBuffer();
			}
		}

		public void printf(String s, Object... o) {
			print(String.format(s, o));
		}

		private void flushToBuffer() {
			try {
				os.write(sb.toString().getBytes());
			}catch(IOException e) {
				e.printStackTrace();
			}
			sb = new StringBuilder(BUFFER_SIZE);
		}

		public void flush() {
			try {
				flushToBuffer();
				os.flush();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

		public void close() {
			flush();
			try {
				os.close();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

	}

	static class Utilities {
		public static class Debug {
			public static final boolean LOCAL = System.getProperty("ONLINE_JUDGE")==null;

			private static <T> String ts(T t) {
				if(t==null) {
					return "null";
				}
				try {
					return ts((Iterable) t);
				}catch(ClassCastException e) {
					if(t instanceof int[]) {
						String s = Arrays.toString((int[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}else if(t instanceof long[]) {
						String s = Arrays.toString((long[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}else if(t instanceof char[]) {
						String s = Arrays.toString((char[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}else if(t instanceof double[]) {
						String s = Arrays.toString((double[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}else if(t instanceof boolean[]) {
						String s = Arrays.toString((boolean[]) t);
						return "{"+s.substring(1, s.length()-1)+"}";
					}
					try {
						return ts((Object[]) t);
					}catch(ClassCastException e1) {
						return t.toString();
					}
				}
			}

			private static <T> String ts(T[] arr) {
				StringBuilder ret = new StringBuilder();
				ret.append("{");
				boolean first = true;
				for(T t: arr) {
					if(!first) {
						ret.append(", ");
					}
					first = false;
					ret.append(ts(t));
				}
				ret.append("}");
				return ret.toString();
			}

			private static <T> String ts(Iterable<T> iter) {
				StringBuilder ret = new StringBuilder();
				ret.append("{");
				boolean first = true;
				for(T t: iter) {
					if(!first) {
						ret.append(", ");
					}
					first = false;
					ret.append(ts(t));
				}
				ret.append("}");
				return ret.toString();
			}

			public static void dbg(Object... o) {
				if(LOCAL) {
					System.err.print("Line #"+Thread.currentThread().getStackTrace()[2].getLineNumber()+": [");
					for(int i = 0; i<o.length; i++) {
						if(i!=0) {
							System.err.print(", ");
						}
						System.err.print(ts(o[i]));
					}
					System.err.println("]");
				}
			}

		}

	}
}