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 <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;

#define endl '\n'
#define fi first
#define se second
#define pb push_back
#define len(a) a.size()
#define pll pair<ll,ll>
#define pii pair<int,int>
#define mp(x, y) make_pair((x), (y))
#define all(con) con.begin(),con.end()
#define sa(x) sort((x).begin(), (x).end())
#define show(x) cout<< #x <<" is "<< x <<"\n"

ll mod=1e9+7;

/* (x^y)%p */
ll po(ll x, ll y, ll p) 
{ 
    ll res = 1; 
    x = x % p; 
    if (x == 0) return 0; 
    while (y > 0) 
    { 
        if (y & 1) 
            res = (res*x) % p; 
        y = y>>1;  
        x = (x*x) % p; 
    } 
    return res; 
} 

void solve() {
    ll n,ki;
    cin>>n>>ki;
    double po=0,cu=1;
    ll A[n];
    for(int i=0;i<n;i++)
        cin>>A[i];
    ll k=n;
    while(A[k-1]==k and k>0)
        k--;
    for(int i=0;i<ki;i++) {
        ll z;
        double x;
        cin>>z>>x;
        if(z>=k) {
            po+=cu*x;
            cu*=(1-x);
            // cout<<cu<<endl;
        }
    }
    if(k==0)
        cout<<fixed<<setprecision(6)<<1.000000<<endl;
    else
        cout<<fixed<<setprecision(6)<<po<<endl;
}

int main() {
    ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    int t=1;
    cin>>t;
    while(t--) {
        solve();
    }
    return 0;
}