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> 
#define ll long long
using namespace std;
#define inf 1000000000000000005
#define mod 1000000007
#define pi 3.1415926536
#define all(c) (c).begin(), (c).end()
ll fpow(ll n, ll k, ll p = mod) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n%p; n = n * n%p;} return r;}
ll inv(ll a, ll p = mod) {return fpow(a, p - 2, p);}

int n,m,arr[100009],see[100009];



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(int i=1;i<=n;i++){
			cin >> arr[i];
			see[i] = 0;
		}
		for(int i=0;i<100009;i++){
			see[i] = 1;
		}
		if(arr[n]!=n){
			see[n] = 0;
		}
		for(int i=n-1;i>=1;i--){
			if(see[i+1]==1){
				if(arr[i]==i){
					see[i] = 1;
				}
				else{
					see[i] = 0;
				}
			}
			else{
				see[i] = 0;
			}
		}
		vector<double> v1;
		vector<double> v2;
		for(int i=0;i<m;i++){
			int nn;
			double kk;
			cin >> nn>>kk;
			if(see[nn+1]){
				v1.push_back(kk);
				if(v2.size()>0){
					double t1 = v2.back();
					double t2 = double((double)1.0-kk);
					t2 = t2*t1;
					v2.push_back(t2);
				}
				else{
					double t2 = double((double)1.0-kk);
					v2.push_back(t2);
				}
			}
		}
		if(see[1]){
			double ans = 1.0;
			cout<<fixed;
			cout<<setprecision(6);
			cout<<ans<<"\n";
			continue;
		}
			double ans = 0.0;
			if(v1.size()>0){
				ans+=v1[0];
			}
			for(int i=1;i<v1.size();i++){
				double kk = v2[i-1];
				kk *= v1[i];
				ans+=kk;
			}
			cout<<fixed;
			cout<<setprecision(6);
			cout<<ans<<"\n";
	}
}