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 <cstdio>
#include <queue>
#include <vector>
#include<iostream>
#include<map>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<bits/stdc++.h>

#define rep(i,a,b)  for(register  int i(a); i <= b; ++i)
#define ll long long
#define inf 1e15
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define PI acos(-1)
#define fi first
#define se second
#define pb push_back
#define debug(x)   printf("x=%d\n",x);
#define ls(x) arr[x].child[0]
#define rs(x) arr[x].child[1]
using namespace std;

const int N=1e6+10;
const ll mod=1e9+7;

inline ll read()
{
    ll X=0; bool flag=1; char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') flag=0; ch=getchar();}
    while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
    if(flag) return X;
    return ~(X-1);
}


ll n,m,k;
int ar[N],lz[N];
ll ans;
vector<pair<int,double> >vt;
void solvve()
{
    n=read(); m=read(); vt.clear();
    rep(i,1,n)
    ar[i]=read();
    rep(i,1,m)
    {
        int k1=read(); double p; cin>>p;

        pair<int,double>t; t.fi=k1; t.se=p;
        vt.push_back(t);
    }
    sort(vt.begin(),vt.end()); reverse(vt.begin(),vt.end());
    int lst=n+1;
    for(int i=n;i>=1;i--)
    {
        if(ar[i]!=i) break;
        lst=i;
    }
    lst--;
    double ans=1.0;
    double tp=1.0;
    for(auto it:vt)
    {
        if(it.fi>=lst) tp*=(1.0-it.se);
    }
    ans-=tp;
    if(lst==0) ans=1.0;
    printf("%.10f\n",ans);
}
int main()
{
    // gbtb;
   //  freopen("input","r",stdin);

      //  while(scanf("%lld",&n)!=EOF)

       int T;T=read();while(T--)
    {
        //printf("cas=%d   ",++cc);
        solvve();
    }
    return 0;

}