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).

<image>

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;
typedef long l;
typedef long long ll;
# define MOD 1000000007
# define PI 3.14
#define vi vector<int>
#define vl vector<l>
#define vll vector<ll>
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define ml map<l,l>
#define mll map<ll,ll>
#define mcl map<char,ll>
# define umll unorderd_map<ll,ll>
# define umcl unordered_map<char,l>
#define pll pair<ll,ll>
# define forl(i,range) for(l i=0;i<range;i++)
# define forll(i,range) for(ll i=0;i<range;i++)
// int ans=upper_bound(start,end ,value)-start;
// auto itr=uper_bound(start,end ,value)-vector.begin();
ll hcf(ll x, ll y) {	//returns gcd of two long long ints
	if (x == 0) return y;
	return hcf(y % x, x);
}
ll lcm(ll a,ll b){
	return((a*b)/hcf(a,b));
 
}
vll all_prime(ll n)
{   vll sieve(n+1,0);
    for (ll x = 2; x <= n; x++) 
    {
        if (sieve[x]) continue;
        for (int u = 2*x; u <= n; u += x)
        {
            if(sieve[u]==0)
            {
                sieve[u]=x;
            }
        }
    }
        return sieve;
}
ll power(ll x,ll n)
{
    if(n==0)return 1;
    ll temp=power(x,n/2);
    if(n%2==0) return temp*temp;
    else return x*temp*temp;
}
 
 
 
/*  ************************************************************************************************************** */
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
     ll t;
     cin>>t;
     while(t--){
         ll n,m,inc=0;
         double prod=1;
         cin>>n>>m;
         ll sum=n;
         ll a[n];
         forll(i,n){cin>>a[i];}
         ll r[m];
         double p[m];
         forll(i,m)
         {cin>>r[i]>>p[i];}
         for(ll i=n-1;i>-1;i--)
         {
             if(i+1==a[i]){
                 inc++;
                 sum=i;
             }else break;
         }
         if(inc==n) 
         {inc=1;cout<<fixed<<setprecision(6)<<float(inc)<<"\n";}
         else {
            forll(i,m)
             { if(r[i]>=sum){ prod=prod*(1-p[i]);}}
             cout<<fixed<<setprecision(6)<<1-prod<<"\n";
             }
         }
    }