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<stdio.h>
#include<string.h>
#include<map>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<stack>
#include<math.h>
#include<utility>
#include<set>
using namespace std;
#define fi first
#define sc second
#define ll long long
#define pb push_back
#define INF 0x3f3f3f3f
#define ls 2*tr
#define rs 2*tr+1
#define N 102000

inline ll re()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){ x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
    return x*f;
}
int a[N];
int dp[N];
struct node
{
    int l;
    double r;
}e[N];
bool cmp(node l,node r)
{
    return l.l<r.l;
}
int main()
{
    int t=re();
    while(t--)
    {
        int n=re(),m=re();
        for(int i=1;i<=n;i++) dp[i]=0;
        for(int i=1;i<=n;i++) a[i]=re();
        dp[n+1]=1;
        for(int i=n;i>=1;i--)
        {
            if(a[i]==i) dp[i]=dp[i+1];
        }
        double ans=0;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%lf",&e[i].l,&e[i].r);
        }
        sort(e+1,e+1+m,cmp);
        double p=1;
        for(int i=m;i>=1;i--)
        {
            if(dp[e[i].l+1])
            {
                ans+=p*e[i].r;
                p*=(1-e[i].r);
            }
            else
            {
                break;
            }
        }
        if(dp[1]) ans+=p;
        printf("%.6lf\n",ans);
    }
}