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;

void solve() {

    int n,m;
    cin>>n>>m;
    int i;
    vector<int> arr;
    for(i=0;i<n;i++)
    {
        int temp;
        cin>>temp;
        arr.push_back(temp);
    }
    bool flag = false;
    if(is_sorted(arr.begin(), arr.end()))
    {
        flag = true;
    }
    vector<bool> sorted;
    sorted.resize(n+1);
    fill(sorted.begin(), sorted.end(), false);
    sorted[n] = true;
    for(i=n-1;i>=0;i--)
    {
        if(arr[i] == i+1)
            sorted[i] = true;       
        else
        {
            break;
        }
    }
    vector<double> ans;
    for(i=0;i<m;i++)
    {
        int r;
        double p;
        cin>>r>>p;
        if(r!=1)
        {
            if(sorted[r] == true)
                ans.push_back(p);
        }
    }
    if(flag)
    {
        cout << 1.000000 << endl;
        return;
    }
    double final_ans = 0.0;
    vector<double> dp;
    dp.resize(ans.size()+1);
    dp[0] = 1.0;
    for(i=1;i<ans.size();i++)
    {
        dp[i] = dp[i-1]*(1-ans[i-1]);
    }
    for(i=0;i<ans.size();i++)
    {
        double term = dp[i];
        // for(int j=0;j<i;j++)
        // {
        //     term = term*(1-ans[j]);
        // }
        term = term*ans[i];
        final_ans+= term;
    }
    cout << final_ans << endl;

}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    int tc; cin >> tc;
    for (int t = 1; t <= tc; t++) {
        // cout << "Case #" << t  << ": ";
        solve();
    }
}