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 ll long long int
#define ull unsigned long long int
#define fasterInOut ios::sync_with_stdio(false); cin.tie(0);
#define pi acos(-1)
 
int main()    {
//	freopen("/home/mimi0007/Desktop/Temp/in", "r", stdin);
	fasterInOut;
	ll T;
	cin >> T;
	while(T--) {
		ll n, m;
		cin >> n >> m;
		ll c = 0, l;
		for(int i = 1; i <= n; i++) {
			cin >> l;
			if(l != i) {
				c = i;
			}
		}
		
		ll a;
		double b;
		vector <double> v;
		while(m--) {
			cin >> a >> b;
			if(a >= c) {
				v.push_back(b);
			}
		}
		
		if(c == 0) {
			cout << "1.000000" << endl;
			continue;
		}
		
		else {
			if(v.size() == 1) {
				cout << setprecision(6) << fixed << v[0] << endl;
			}
			
			else if(v.size() == 0) {
				cout << "0.000000" << endl;
			}
			
			else {
				double pa = v[0], p;
				for(int i = 1; i < v.size(); i++) {
					p = (pa+v[i]) - (pa*v[i]);
					pa = p;
				}
					
				cout << setprecision(6) << fixed << p << endl;
			}
		}
	}	
		
	return 0;
}