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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
#include<set>
const int mod = 1000000007;
using namespace std;
bool isPalindrome(string str)
{
    // Start from leftmost and rightmost corners of str
    int l = 0;
    int h = str.length() - 1;

    // Keep comparing characters while they are same
    while (h > l)
    {
        if (str[l++] != str[h--])
        {
            return false;
        }
    }
    return true;
}
long long fact[300500];
long long invFact[300500];

long long fast_pow(long long a, long long p) {
    long long res = 1;
    while (p) {
        if (p % 2 == 0) {
            a = (a * a) % mod;
            p /= 2;
        } else {
            res = (res * a) % mod;
            p--;
        }
    }
    return res;
}
long long C(int n,int k)
{
    if (k > n) {
        return 0;
    }
    return fact[n] * invFact[k] % mod * invFact[n - k] % mod;
}
void solve()
{
  long long n,m=0,k=0;
  cin>>n>>m;
  vector<int> a(n);
  vector<pair<int,float>> dp(m);
  for(int i=0;i<n;i++)
    cin>>a[i];
    for(int i=0;i<m;i++)
    {
        int x;float y;
        cin>>x>>y;
        dp[i]={x,y};
    }
   int i=n-1;
   while(i>=0 && a[i]==i+1)
    i--;
   if(i==-1)
   {
       cout<<1.00<<endl;
       return ;
   }
   int var=i;float succ=0.0, uns=0.0;
   for(int i=0;i<m;i++)
   {
       int x=dp[i].first-1;
       if(x<var)
        continue;
        //cout<<x<<"a"<<endl;
        if(succ==0.0)
        {
            succ=dp[i].second;
            uns=1-dp[i].second;
        }
        else
        {
            succ+=uns*dp[i].second;
            uns=uns*(1-dp[i].second);
        }

   }
   cout<<succ<<endl;

}
int main()
{
    	    ios_base::sync_with_stdio(false); cin.tie(0);
#ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
#endif
    /**fact[0] = invFact[0] = 1;
    for (int i = 1; i < 300500; i++) {
        fact[i] = (fact[i - 1] * i) % mod;
        invFact[i] = fast_pow(fact[i], mod - 2);
    }**/
	int t;
	cin>>t;
	while(t--){
          solve();
	}
}