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 CF1461C{
  static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
        
    } 
    public static class Pair implements Comparable<Pair>{
        int vtx; int val;
        public Pair(int a, int b){
            this.vtx=a; this.val=b;
        }
        public int compareTo(Pair other){
            if(this.val<other.val)return -1;
            if(this.val>other.val)return 1;
            if(this.vtx<other.vtx)return -1;
            return 1;
        }
    }
    static int MOD=998244353;
    static int[] rk, p,siz;
    public static void main(String[] args){
        FastReader br=new FastReader();
        int T=br.nextInt(); int[]p; int x; int r; double pr;
        while(T>0){
            T--;int N=br.nextInt(); int M=br.nextInt();
            p=new int[N];
            for (int i = 0; i < N; i++) {
                p[i]=br.nextInt()-1;
            }
            x=N-1;
            while(x>=0 && p[x]==x){x--;}
            
            double fail=Math.min(x+1,1);
            for (int i = 0; i < M; i++) {
                r=br.nextInt()-1; pr=1-br.nextDouble();
                if(r>=x){
                    fail*=pr;
                }
            }
            System.out.println(1-fail);
        }

    }
    
    public static int find(int x, int[] p){
        if(p[x]==x)return x;
        int ans=find(p[x],p); p[x]=ans; return ans;
    }
    
    public static void merge(int a, int b) {
        if(rk[a]<rk[b]) {
            p[a]=b; siz[b]+=siz[a];
        }else if(rk[a]==rk[b]) {
            p[a]=b; rk[b]++;siz[b]+=siz[a];
        }else {
            p[b]=a; siz[a]+=siz[b];
        }
    }
    
}
//Debugging:
//Are you sure your algorithm is correct?
//Bounds: long
//Special cases: n=0,1?
//Make sure you remove your debugging code before you submit!