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:
//Lets Begin With Name Of Lord Mahadev..
#include<bits/stdc++.h>
#define mod 1000000007
#define pi 3.1415926535
#define ll long long int
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define endl "\n"
#define cy cout<<"YES"<<endl;
#define cn cout<<"NO"<<endl;
#define pb(v) push_back(v);
const int MaxN=1e5+5;
using namespace std;

ll t,n,m,k,r,a[MaxN];
ll alreadysorted[MaxN],willbesorted[MaxN],sortedtillhere[MaxN];
string s;

int main(){
    fast;
    cin>>t;

    while(t--){
        cin>>n>>m;

        for(int i=0;i<n;i++){
            cin>>a[i];
            if(i == 0){
                alreadysorted[i]=1;
                sortedtillhere[i]=1;
            }else{
                if(alreadysorted[i-1] && a[i] > a[i-1]){
                    alreadysorted[i] = 1;
                }else{
                    alreadysorted[i] = 0;
                }
            }
            willbesorted[i]=0;
        }
        ll x=n;
        willbesorted[n]=1;
        for(int i=n-1;i>=0;i--){
            if(a[i] == x){
                willbesorted[i]=1;
            }else{
                break;
            }
            x--;
        }
//        for(int i=0;i<n;i++){
//            cout<<alreadysorted[i]<<" "<<willbesorted[i]<<endl;
//        }
        double ans1=0,ans2=0,ans,p;
        for(int i=0;i<m;i++){
            cin>>r>>p;
            r--;
            if(alreadysorted[r]){
                //cout<<"skipped"<<endl;
                continue;
            }
            if(willbesorted[r+1]){
//                ans1=(ans1+ans2)*p;
//                ans2=ans2*(1.0-p);
//                if(ans1 == 0){
//                    ans1=p;
//                }
//                if(ans2 == 0){
//                    ans2=1.0-p;
//                }
                ans1=ans1+ans2*p;

                ans2=ans2*(1-p);
                if(ans1 == 0){
                    ans1=p;
                }
                if(ans2 == 0){
                    ans2=1.0-p;
                }
                ans=ans1;
                //cout<<ans1<<" "<<ans2<<endl;
            }else{

            }
        }
        if(alreadysorted[n-1] || ans1>1.0){
            ans1=1;
        }
        cout<<fixed<<setprecision(10)<<ans1<<endl;
    }
    return 0;
}