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:
/*
#pragma GCC optimize("O2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,fma")
//*/
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int  ,int > pii;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll maxn  = 3e6;
const ll mod =1e9+7;
const ld PI = acos((ld)-1);

#define pb push_back
#define endl '\n'
#define dokme(x) cout << x , exit(0)
#define migmig ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define ms(x , y) memset(x , y , sizeof x)
ll pw(ll a, ll b, ll md = mod){ll res = 1;while(b){if(b&1){res=(a*res)%md;}a=(a*a)%md;b>>=1;}return(res);}

int q;
int n , m;
int a[maxn];

ld p[maxn];
int r[maxn];

int32_t main(){
    migmig;
	cin >> q;
	while(q --){
		cin >> n >> m;
		for(int i = 1 ; i <= n ; i ++)
			cin >> a[i];
		for(int i = 1 ; i <= m ; i ++)
			cin >> r[i] >> p[i];
		int lst = -1;
		for(int i = n ; i ; i --){
			if(a[i] != i){
				lst = i;
				break;
			}
		}
		if(lst == -1){
			cout << 1 << endl;
			continue;
		}
		ld ans = 1;
		for(int i = 1 ; i <= m ; i ++){
			if(r[i] >= lst)ans*=(1 - p[i]);
		}
		cout << fixed << setprecision(10) << 1 - ans << endl;
	}



    return(0);
}