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.math.*;
import java.util.*;

public class CF689c {
	static PrintWriter pw;

	static class FastReader {
		private InputStream mIs;
		private byte[] buf = new byte[1024];
		private int curChar, numChars;

		public FastReader() {
			this(System.in);
		}

		public FastReader(InputStream is) {
			mIs = is;
		}

		public int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = mIs.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public String next() {
			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 long l() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long 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 int i() {
			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 double d() throws IOException {
			return Double.parseDouble(next());
		}

		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

		public void scanIntArr(int[] arr) {
			for (int li = 0; li < arr.length; ++li)
				arr[li] = i();
		}

		public void scanIntIndexArr(int[] arr) {
			for (int li = 0; li < arr.length; ++li) {
				arr[li] = i() - 1;
			}
		}

		public void printIntArr(int[] arr) {
			for (int li = 0; li < arr.length; ++li)
				pw.print(arr[li] + " ");
			pw.println();
		}

		public void printLongArr(long[] arr) {
			for (int li = 0; li < arr.length; ++li)
				pw.print(arr[li] + " ");
			pw.println();
		}

		public void scanLongArr(long[] arr) {
			for (int i = 0; i < arr.length; ++i) {
				arr[i] = l();
			}
		}

		public void shuffle(int[] arr) {
			for (int i = arr.length; i > 0; --i) {
				int r = (int) (Math.random() * i);
				int temp = arr[i - 1];
				arr[i - 1] = arr[r];
				arr[r] = temp;
			}
		}

		public int findMax(int[] arr) {
			return Arrays.stream(arr).max().getAsInt();
		}
	}

	static double fun(int id,int m,int larr[],double parr[],int thr,double dp[][],boolean flg) {
		if(id==m-1) {
			if(flg) {
				if(larr[id]>=thr)
					return parr[id];
				return 0;
			}
			return 0;
		}
		if(flg && dp[id][0]!=0)
			return dp[id][0];
		if(!flg && dp[id][1]!=0)
			return dp[id][1];
		
		if(flg) {
			if(larr[id]>=thr) {
				dp[id][0]=parr[id];
				return dp[id][0];
			}
			else {
				dp[id][0] = parr[id]*(fun(id+1,m,larr,parr,thr,dp,true)+fun(id+1,m,larr,parr,thr,dp,false));
			}
			return dp[id][0];
		}
		
		
		dp[id][1] = (1-parr[id])*(fun(id+1,m,larr,parr,thr,dp,true)+fun(id+1,m,larr,parr,thr,dp,false));
		return dp[id][1];
	}
	
	public static void main(String[] args) throws IOException {
		FastReader fr = new FastReader();
		pw = new PrintWriter(System.out);
		int i, j, n, m, temp, t, k;
		t = fr.i();
		while (t-- > 0) {
			n = fr.i();
			m = fr.i();
			int arr[] = new int[n];
			int arr1[] = new int[n];
			for(i=0;i<n;i++) {
				arr[i]=fr.i();
				arr1[i]=arr[i];
			}
			Arrays.sort(arr1);
			int thr = n-1;
			while(thr>=0 && arr1[thr]==arr[thr])
				--thr;
			++thr;
			double ans=0;
			double dp[][] = new double[m][2];
			int larr[] = new int[m];
			double parr[] = new double[m];
			for(i=0;i<m;i++) {
				larr[i] = fr.i();
				parr[i] = fr.d();
			}
			ans+=fun(0,m,larr,parr,thr,dp,true);
			ans+=fun(0,m,larr,parr,thr,dp,false);
			if(thr==0)
				ans=1;
			pw.println(ans);
		}
		pw.flush();
		pw.close();
	}
}