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 f1(a,b)   for( long long int i= a; i<=b; i++)
#define f2(a,b)   for( long long int i= a; i<b; i++)
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define ll long long int
#define MOD 1000000007
class Pair{
public:
    ll x;
    double y;
    Pair(ll x,double y)
     : x(x), y(y)
      {
    }
};
bool cmp(Pair a,Pair b)
{
    return a.x<b.x;
}
using namespace std;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
ll n, minlen,m,t,k,x,d1,d2,v1,v2,y;
double prob,ans;
cin>>t;
f1(1,t){
cin>>n>>m;
ll ar[n+1],ck=0;
minlen=-1;
f1(1,n)
{
    cin>>ar[i];
    if(ar[i]!=i)
        minlen=i;
}
vector<pair<ll,double>>v;
f2(0,m)
{
    cin>>k>>prob;
    v.PB(MP(k,prob));
}
sort(v.begin(),v.end());
//f2(0,m)
//{
//    cout<<v[i].F<<" "<<v[i].S<<endl;
//}
cout<<fixed<<setprecision(6);
if(minlen==-1)
    ans=1;
else{


        ans=1;
        f2(0,m)
        {
            if(v[i].F>=minlen){
                    if(v[i].S==1)
                        ans=1;
            else
            ans=ans*(1.0-v[i].S);}
        }
        ans=1-ans;
         ll a=v[m-1].F;
    f1(a+1,n)
    {
        if(ar[i]!=i)
        { ans=0;
        break;}
    }
}
   cout<<ans<<endl;
}
    return 0;
}