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>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
using namespace std;
#define int long long
#define fr(i,a,b) for(int i=a;i<b;i++)
#define frn(i,a,b) for(int i=a;i>=b;i--)
#define vi vector<int>
#define pii pair<int,int>
#define vpi vector<pair<int,int>>
#define vvi vector<vector<int>>
#define test int t;cin>>t; while(t--)
#define mp make_pair
#define pb push_back
#define pu push
#define fi first
#define se second
#define mod 1000000007
/*
typedef tree<int, null_type, less<int>, rb_tree_tag,
        tree_order_statistics_node_update> seti;*/
void fast() {
  ios_base::sync_with_stdio(false); cin.tie(NULL);
#ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin); 
  freopen("output.txt", "w", stdout);
#endif
}


int32_t main(){
  fast();
  test{
    int n,m;
    cin>>n>>m;
    int a[n+1];
    int b[n+1];
    fr(i,0,n) cin>>a[i];
    fr(i,0,n) b[i]=a[i];
    sort(b,b+n);
    int l = -2;
    fr(i,0,n) if(a[i]!=b[i]) l = i;
    long double c;
    int d;
    l++;
    long double k = 1.000000;
    long double r = 1.000000;
    fr(i,0,m){
      cin>>d;
      cin>>c;
      if(d>=l) {
        //cout<<c<<endl;
        k = k*(1.00-c);
      //  cout<<k<<endl;
      }
     // cout<<d<<endl;
    }
    if(l==-1) cout<<"1.000000"<<endl;
    else cout<<fixed<<setprecision(9)<<(1.0-k)<<endl;
  }
  return 0;
}