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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.StringTokenizer;

public class C {

	public static void main(String[] args) throws IOException {

		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = sc.nextInt();
		while(t-- != 0) {
			int n =sc.nextInt(),m = sc.nextInt();
			int arr[] = sc.readArray(n);
			int cc = 0;
			for(int i=n-1; i>=0; i--) {
				if(arr[i] == i+1)cc++;
				else break;
			}
			int req = n - cc;
			ArrayList<Pair> lis = new ArrayList<C.Pair>();
			for(int i=0; i<m; i++) {
				int a = sc.nextInt();
				double p = sc.nextDouble();
				if(a >= req) {
					lis.add(new Pair(a, p));
				}
			}
			if(req == 0) {
				System.out.println(1.0);
				continue;
			}
			m = lis.size();
			double dp[][] = new double[m+1][2];
			
			dp[0][0] = 1;
			for(int i=1; i<=m; i++) {
				// not perform operation
				dp[i][0] = dp[i-1][0]*(1-lis.get(i-1).y);
				dp[i][1] = dp[i-1][0]*(lis.get(i-1).y) + dp[i-1][1];
			}
			System.out.println(dp[m][1]);
		}
		

	}

	static void ruffleSort(int[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			int temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
		Arrays.sort(a);
	}

	static void ruffleSort(long[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			long temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
		Arrays.sort(a);
	}

	static class Pair implements Comparable<Pair> {
		int x;
		double y;

		Pair(int x, double y) {
			this.x = x;
			this.y = y;
		}

		@Override
		public int compareTo(Pair o) {
			return Integer.compare(this.x, o.x);
		}

	}

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

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

		long[] readArrayLong(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	static int ceil(int x, int y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static long ceil(long x, long y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static int max(int x, int y) {
		return Math.max(x, y);
	}

	static int min(int x, int y) {
		return Math.min(x, y);
	}

	static int abs(int x) {
		return Math.abs(x);
	}

	static long abs(long x) {
		return Math.abs(x);
	}

	static int log2(int N) {
		int result = (int) (Math.log(N) / Math.log(2));
		return result;
	}

	static long max(long x, long y) {
		return Math.max(x, y);
	}

	static long min(long x, long y) {
		return Math.min(x, y);
	}

	static int min(int a[]) {
		int x = 1_000_000_00_9;
		for (int i = 0; i < a.length; i++)
			x = min(x, a[i]);
		return x;
	}

	static int max(int a[]) {
		int x = -1_000_000_00_9;
		for (int i = 0; i < a.length; i++)
			x = max(x, a[i]);
		return x;
	}

	static long min(long a[]) {
		long x = (long) 3e18;
		for (int i = 0; i < a.length; i++)
			x = min(x, a[i]);
		return x;
	}

	static long max(long a[]) {
		long x = -(long) 3e18;
		for (int i = 0; i < a.length; i++)
			x = max(x, a[i]);
		return x;
	}

	static int power(int x, int y) {
		int res = 1;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * x);
			y >>= 1;
			x = (x * x);
		}
		return res;
	}

	static long power(long x, long y) {
		long res = 1;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * x);
			y >>= 1;
			x = (x * x);
		}
		return res;
	}

	static void intsort(int[] a) {
		List<Integer> temp = new ArrayList<Integer>();
		for (int i = 0; i < a.length; i++)
			temp.add(a[i]);
		Collections.sort(temp);
		for (int i = 0; i < a.length; i++)
			a[i] = temp.get(i);
	}

	static void longsort(long[] a) {
		List<Long> temp = new ArrayList<Long>();
		for (int i = 0; i < a.length; i++)
			temp.add(a[i]);
		Collections.sort(temp);
		for (int i = 0; i < a.length; i++)
			a[i] = temp.get(i);
	}

	static void reverseintsort(int[] a) {
		List<Integer> temp = new ArrayList<Integer>();
		for (int i = 0; i < a.length; i++)
			temp.add(a[i]);
		Collections.sort(temp);
		Collections.reverseOrder();
		for (int i = 0; i < a.length; i++)
			a[i] = temp.get(i);
	}

	static void reverselongsort(long[] a) {
		List<Long> temp = new ArrayList<Long>();
		for (int i = 0; i < a.length; i++)
			temp.add(a[i]);
		Collections.sort(temp);
		Collections.reverseOrder();
		for (int i = 0; i < a.length; i++)
			a[i] = temp.get(i);
	}

	static class longpair implements Comparable<longpair> {
		long x, y;

		longpair(long x, long y) {
			this.x = x;
			this.y = y;
		}

		public int compareTo(longpair p) {
			return Long.compare(this.x, p.x);
		}
	}

	public static int gcd(int a, int b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.intValue();
	}

	public static long gcd(long a, long b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.longValue();
	}
}