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.util.*;
import java.io.*;
import static java.lang.Math.min;
import static java.lang.Math.max;
public class Solver {
    static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static FastScanner sc = new FastScanner();
    static int MAX = Integer.MAX_VALUE;
    static int MIN = Integer.MIN_VALUE;
    public static void main(String[] args) {
        solve();
        out.close();
    }
    
    static void solve(){
    	int t = sc.nextint();
    	while(t-->0) {
    		int n = sc.nextint(), m = sc.nextint();
    		int[] nums = sc.readint(n);
    		int[] origNums = new int[n];
    		for(int i = 0; i < n; ++i) {
    			origNums[i] = nums[i];
    		}

    		Pair[] arr = new Pair[m];
    		for(int i = 0; i < m; ++i) {
    			int l = sc.nextint();
    			double p = sc.nextdouble();
    			arr[i] = new Pair(l, p);
    		}
    		
    		sc.sort(nums);
    		int rb = n;
    		for(int j = nums.length-1; j >= 0; --j) {
    			if(nums[j] == origNums[j]) {
    				rb = j;
    			}
    			else {
    				break;
    			}
    		}
    		
    		if(rb == 0) {
    			out.println(1);
    			continue;
    		}
    		
    		double prob = 1;
    		for(int i = 0; i < m; ++i) {
    			if(arr[i].r >= rb) {
    				prob *= (1-arr[i].p);
    			}
    		}
    		
    		
    		out.println(1-prob);
    	}
    }
    
    static class Pair{
    	int r;
    	double p;
    	public Pair(int r, double p) {
    		this.r = r;
    		this.p = p;
    	}
    }
    
    static StringTokenizer st;
    static BufferedReader br;
    static Random r = new Random();
    static class FastScanner{
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public String next() {
            while(st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public void sort(long[] a) {
        	for(int i = 0; i < a.length; ++i) {
        		int idx = r.nextInt(a.length);
        		long temp = a[idx];
        		a[idx] = a[i];
        		a[i] = temp;
        	}
        	
        	Arrays.sort(a);
        }
        
        public void sort(int[] a) {
        	for(int i = 0; i < a.length; ++i) {
        		int idx = r.nextInt(a.length);
        		int temp = a[idx];
        		a[idx] = a[i];
        		a[i] = temp;
        	}
        	Arrays.sort(a);
        }
        
        public int[] readint(int k) {
            int[] res = new int[k];
            for(int i = 0; i < k; ++i) {
                res[i] = nextint();
            }
            return res;
        }
        
        public long[] readlong(int k) {
            long[] res = new long[k];
            for(int i = 0; i < k; ++i) {
                res[i] = nextlong();
            }
            return res;
        }
        
        public int nextint() {
            return Integer.parseInt(next());
        }
        
        public long nextlong() {
            return Long.parseLong(next());
        }
        
        public double nextdouble() {
            return Double.parseDouble(next());
        }
        
        public String nextline() {
            String res = "";
            try {
                res = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return res;
        }
    }
}