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 int long long
#define float double
#define itn int
#define f(j,n) for(int i=j;i<n;i++)
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL)
#define endl '\n'
#define Endl '\n'
#define ednl '\n'
#define Ednl '\n'
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define mset(arr, i) memset(arr, i, sizeof(arr))
using namespace std;
 
typedef vector<pair<int, int>> vpi;
typedef vector<vector<int>> vvi;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef map<int,int> mii;
typedef priority_queue<int> pqi;
typedef priority_queue<int, vector<int>, greater<int>> pqim;
 
long long binpow(long long a, long long b) { a %= mod; long long res = 1; while (b > 0){if (b & 1)res = res * a % mod;a = a * a % mod;b >>= 1;}return res;}

int32_t main()
{
    int t1;
    fastio;
    cin>>t1;
    //t1=1;
    while(t1--)
    {
        int n, m;
        cin>>n>>m;
        int a[n];
        f(0,n)cin>>a[i];
        int i;
        for(i=n-1;i>=0;i--)
        {
            if(a[i]!=i+1)
            {
                break;
            }
        }
        i++;
        double prev = 1.0, ans=0.0;
        while(m--)
        {
            int r;
            double p;
            cin>>r>>p;
            if(r>=i)
            {
                ans+=prev*p;
                prev*=((double)1.0-p);
            }
        }
        if(i==0)
        {
            cout<<1.0<<Endl;
        }
        else
        cout<<ans<<endl;
    }
}