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 <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <cmath>
#include <climits>
#include <algorithm>
#include <map>
#include <stack>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <utility>
#include <iomanip>

using namespace std;
typedef long long int ll;
ll pw(ll a, ll b, ll m){
	ll ans=1;
	while(b>0){
		if(b&1) ans=(ans*a)%m;
		a=(a*a)%m;
		b>>=1;
	}
	return ans;
}

ll gcd(ll a,ll b){
	if(a==0) return b;
	return gcd(b % a , a);
}

bool comp(const pair<int,int> &a,const pair<int,int> &b){
    return a.second>b.second;
}

ll n,m;
set<ll> has;
vector<ll> input;

void solve(ll l,ll r){
	ll sum=0;
	for(ll i=l;i<r+1;i++){
		sum+=input[i];
	}
	has.insert(sum);
	ll mid=(input[l]+input[r])/2;
	ll pos=-1;
	for(ll i=l;i<r+1;i++){
		if(input[i]<=mid){
			pos=i;
		}
		else{
			break;
		}
	}
	if(pos==-1 || pos==r) return;
	solve(l,pos);
	solve(pos+1,r);
}
void randomevent(){
	ll right=n;
	while(right>=0 && input[right-1]==right){
		right--;
	}
	double f=(right>0) ? 1:0;
	for(ll i=0;i<m;i++){
		ll r;cin>>r;
		double p;cin>>p;
		if(r>=right){
			f*=(1-p);
		}
	}
	cout<<fixed<<setprecision(12)<<(1-f)<<endl;
}



int main(){
#ifndef ONLINE_JUDGE
    freopen("input.txt","r", stdin);
    freopen("output.txt","w", stdout);
#endif 

    int t;cin>>t;
    while(t--){
    	cin>>n>>m;
    	for(ll i=0;i<n;i++){
    		ll p;cin>>p;
    		input.push_back(p);
    	}
    	randomevent();
    	input.clear();

    }
}