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;
#define f(i,b) for(int i=1;i<=b;++i)
#define pb push_back
#define pa pair<int,int>
#define F first 
#define S second
#define mp make_pair
#define ve vector<int>
#define v1 vector<string>
#define ma map<int,int>
#define se set<int>
#define ll long long
#define B begin()
#define E end()
#define pu push
#define ff(i,n) for(int i=0;i<n;++i)
#define fo(i,n) for(int i=n-1;i>=0;--i)
#define lb lower_bound
#define ub upper_bound
#define pq priority_queue<N,vector<N>,co>

int ch(ve &v,ve a,int n,int i,ve &an){
	if(i==n-1){
		v[i]=1;
		an[i]=1;
		return 1;
	}
	if(an[i]!=-1)
		return an[i];
	an[i] = ch(v,a,n,i+1,an) && a[i+1]==i+2;
	return an[i];
}
void solve(){
	int n,m,r=1;
	cin>>n>>m;
	ve a;
	f(i,n){
		int x;
		cin>>x;
		if(x!=i)
			r=0;
		a.pb(x);
	}

	ve v,an;
	f(i,n){
		v.pb(0);
		an.pb(0);
	}
	an[n-1]=1;
	//ch(v,a,n,0,an);
	for(int i=n-2;i>=0;--i){
		if(an[i+1]==1 && a[i+1]==i+2)
			an[i]=1;
	}
	double pr=0.0;
	f(i,m){
		int b;
		double p;
		cin>>b>>p;
		//cout<<an1[b]<<endl;
		if(an[b-1])
			pr=(1-pr)*p+(1-p)*pr+p*pr;
	}
	if(r)
		cout<<"1.000000\n";
	else
		cout<<pr<<endl;
}
int main(){
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL);
	int t=1;
	cin>>t;
	while(t--)
		solve();
	
	return 0;
}