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:
//karne se hi hota hai ;)
#include <bits/stdc++.h>
#define int long long int
#define mod 1000000007
#define endl '\n'
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#define decimal(x) cout<<fixed<<setprecision(x);
bool prime[100005]; 
void Sieve(int n) {
    memset(prime, true, sizeof(prime)); 
  
    for (int p=2; p*p<=n; p++) 
    { 
        if (prime[p] == true) 
        { 
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
  
} 
const int N=1e5+5;
 
double dp[N][2];
void solve()
{
    decimal(9)
    int n,m;
    cin>>n>>m;
    int arr[n];
    for(int i=0;i<n;i++)cin>>arr[i];
    int cnt=0;
    for(int i=n-1;i>=0;i--)
    {
        if(arr[i]==i+1)cnt++;
        else break;
    }
    int req=n-cnt;
    vector<double>p;
    for(int i=0;i<m;i++)
    {
        int a;double b;
        cin>>a>>b;
        if(a>=req)p.push_back(b);
    }
    if(req==0)cout<<"1.0"<<endl;
    else {
        int k=p.size();
        dp[0][0]=1;
        for(int i=1;i<=k;i++)
        {
            dp[i][0]=dp[i-1][0]*(1-p[i-1]);
            dp[i][1]=dp[i-1][1]+dp[i-1][0]*(p[i-1]);
        }
        cout<<dp[k][1]<<endl;
    }
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t=1;
    cin>>t;
    while(t--)
    {
        solve();
    }
}