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:
                                                                   //author rhythmicankur
                                                                  //god_is_almighty
#pragma GCC optimize("O3")
#pragma GCC target ("sse4")
#include<bits/stdc++.h>
#define f 		first
#define s 		second
#define int     long long
#define ll      int
#define ld 		long double
#define pb 		push_back
#define eb 		emplace_back
#define mk 		make_pair
#define vi 		vector<int>
#define pii 		pair<int,int>
#define pip             pair<int,pii>
#define vpi 		vector<pii>
#define sbcount(x) __builtin_popcountll(x)
#define REP(i,n) for (int i = 1; i <= n; i++)
#define trav(a, x)      for(auto& a : x)
#define all(x)          x.begin(), x.end()
#define fill(a,val)	memset(a,val,sizeof(a))
#define 	PI acos(-1)
#define coud(a,b) cout<<fixed << setprecision((b)) << (a)
#define M1 	998244353
#define M2 	1000000007
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

template<typename T> T gcd(T a,T b) { if(a==0) return b; return gcd(b%a,a); } 
#define test4(x,y,z,a) 		    cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<"  |  "<<#a<<": "<<a<<endl;
#define test1(x)                cerr<<#x<<": "<<x<<endl
#define test2(x, y)             cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define test3(x, y, z)          cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
ll power(ll a,ll b,ll m=M2){ll answer=1;while(b){if(b&1)answer=(answer*a)%m;b/=2;a=(a*a)%m;}return answer;}
using namespace std;
void ctrl(){cout<<"Control"<<endl;}
int make_num(string p){stringstream geek(p); int x = 0; geek >> x;return x;}
string make_str(int x){ostringstream str1; str1 << x; string geek = str1.str(); return geek;}

signed main()
{	ios::sync_with_stdio(0);
	cin.tie(0); 
	cin.exceptions(cin.failbit);
	int tc;
	cin>>tc;
	for(int t=1;t<=tc;t++)
	{
        int n,m;
        cin>>n>>m;
        int A[n];
        for(int i=0;i<n;i++)
        cin>>A[i];
        int last=0;
        for(int i=n-1;i>=0;i--)
        {    
            if(A[i]!=i+1)
            {
                    last=i+1;
                    break;
            }
        }
        
        vector<double>v;
        double val=1.0;
        for(int i=0;i<m;i++)
        {   
            int r;
            double d;
            cin>>r>>d;
            if(r>=last)
            val = val * (1.0-d);

        }
        if(last==0)
        {    
            val = 0.0;
        }
        coud(1.0-val,12);
        cout<<'\n';
	}

return 0;
}