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:
/*
Author : Vidish Joshi
*/

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

//#include<boost/multiprecision/cpp_int.hpp>
//using namespace boost::multiprecision;

#define ll long long int
#define FIO ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define mod 1000000007
#define FOR1(i, m, n) for(i=m; i<n; i++)
#define FOR2(i, n, m) for(i=n-1; i>=m; i--)
#define re(v) ll v; cin>>v
#define mod 1000000007
#define mod2 998244353
#define mp make_pair
#define pb push_back
#define fill(a) memset(a, 0, sizeof(a))
#define start ll t; cin>>t; while(t--)
#define start1 ll t = 1; while(t--)
#define F first
#define S second
#define br "\n"

void solve(){	
	int n, m; cin>>n>>m;
	int a[n], b[n];
	for(int i=0; i<n; i++) cin>>a[i], b[i] = a[i];
	
	int place = -1;
	sort(b, b+n);
	for(int i=n-1; i>=0; --i){
		if(a[i]!=b[i]){
			place = i;
			break;
		}
	}
//	cout<<place<<" bif\n";
	vector<pair<int, double> > v;
	int x; double y;
	
	for(int i=0; i<m; i++){
		cin>>x>>y;
		x--;
		if(x >= place){
			v.pb({x, y});
		}
	}
//	cout<<v.size()<<" ll\n";
	double last = 1;
	double ans = 0;
	for(int i=0; i<v.size(); i++){
		ans += (v[i].S * last);
		last *= ((double(1)) - v[i].S);
	}
	
	if(place == -1){
		cout<<"1.000000000\n";
		return;
	}
	
	cout<<fixed<<setprecision(9)<<ans<<br;
}

int main(){
    FIO;
    start{
        solve();
    }
    return 0;
}