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:
/*
	Solution by:- 
	--------------Sannidhay Vashal
	----------------NIT SRINAGAR
 
*/
#include<iostream>
#include<vector>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<math.h>
#include<climits>
#include<set>
#include<cstring>
#include<unordered_map>
#include <cstdlib> 
#include<cmath>
#include<string>
#include<iomanip>
 
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ll            long long int
#define ull           unsigned long long int
#define cinll(x)      ll x;cin>>x;
#define cini(x)       int x;cin>>x;
#define cins(x)       string x;cin>>x;
#define vect(x)       vector<ll> x;
#define vect1(x)      vector<ll> x;x.push_back(0);
#define pb(x)         push_back(x)
#define mp(x,y)       make_pair(x,y)
#define MAX           1e18
#define MIN           -1000000007
#define MOD           1000000007

using namespace std;

struct custom_compare1{
  bool operator()(pair<ll,ll> a,pair<ll,ll> b){
    if(a.second==b.second){
      return a.first>b.first;
    }
    return a.second<b.second;
  }
};
struct custom_compare2{
  bool operator()(pair<ll,ll> a,pair<ll,ll> b){
    return a.first>b.first;
  }
};

int main(){
  fio;
  cinll(t);
  while(t--){
    cinll(n);cinll(m);
    vector<ll> arr;
    ll pos = n+1;
    for(ll i=0;i<n;i++){
      cinll(x);
      arr.pb(x);
    }
    for(ll i=n-1;i>=0;i--){
      if(i+1==arr[i]){
        pos = i+1;
      }else{
        break;
      }
    }
    pos--;
    long double val = 1;
    while(m>0){
      cinll(ind);
      double p;cin>>p;
      if(ind>=pos){
        val *= (1-p);
      }
      m--;
    }
    if(pos==0)  cout<<setprecision(6)<<fixed<<1.00<<"\n";
    else  cout<<setprecision(6)<<fixed<<1-val<<"\n";
  }
  return 0;
}