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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main() {
	// your code goes here
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	int t,i,j,n,m;
	double p,a,b,ans,u;
	cin>>t;
	while(t--)
	{
	    cin>>n>>m;
	    int A[n];
	    int B[n];
	    int C[n];
	    C[0]=1;
	    B[0]=0;
	    for(i=0;i<n;i++)
	    {
	        cin>>A[i];
	        if(i)
    	    {
    	        if(A[i]>A[i-1])
    	        C[i]=C[i-1]+1;
    	        else
    	        C[i]=1;
    	        
    	        if(A[i]==(i+1))
    	        B[i]=B[i-1]+1;
    	        else
    	        B[i]=0;
    	    }
    	    
    	    if(i==0&&A[i]==(i+1))
    	    B[i]=1;
    	    
	       // cout<<i<<" : "<<B[i]<<endl;
	    }
        ans=0;u=1;
        for(i=0;i<m;i++)
        {
            cin>>j>>p;
            if(C[j-1]==j)
            {
                a=1;
                b=0;
            }
            else
            {
                a=p;
                b=1-p;
            }
            
            if(B[n-1]>=(n-j))
            {
                
                ans+=(a*u);
                u*=b;
                //cout<<ans<<" : "<<u<<endl;
            }
            
        }
	    
	    cout<<ans<<endl;
	    //cout<<endl;
	    
	}
	
	return 0;
}