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>

#define FOR(i,a,b) for(int i = a ; i <= b; i++)
#define FORD(i,a,b) for(int i = a; i >= b; i--)
#define fi first
#define se second
#define mp make_pair
#define ll long long
#define mod 1000000007

using namespace std;

typedef pair<int,int> ii;
typedef pair<int,ii> iii;
typedef pair<ii,ii> iiii;

const int N = 101005;

long long POW(ll so, ll k)
{
    if(k == 0) return 1;
    if(k == 1) return so;
    ll tam = POW(so,k/2);
    if(k % 2 == 0) return so * so % mod;
    else return tam * tam % mod * so % mod;
}
int n,m,a[N],kt[N];

int main()
{
    if(fopen("text.inp","r"))
    {
        freopen("text.inp","r",stdin);
        freopen("text.out","w",stdout);
    }
    int q;
    cin >> q;
    while(q--)
    {
        cin >> n >> m;
        FOR(i,1,n) cin >> a[i];
        if(is_sorted(a+1,a+n+1) == true )
        {
            cout << fixed << setprecision(6) << 1.0 << endl;
            int xx;
            long double yy;
            FOR(i,1,m) cin >> xx >> yy;
            continue;
        }
        long double tich = 1;
        FOR(i,0,n+1) kt[i] = 1;
        FORD(i,n,1) if(a[i] == i) kt[i] = kt[i+1];
            else kt[i] = 0;
        int xx;
        long double yy;
        FOR(i,1,m)
        {
            cin >> xx >> yy;
            //cout << kt[xx+1] << endl;
            if((is_sorted(a+xx+2,a+n+1)==true || xx + 1 > n) && kt[xx+1] == 1)
            {
                tich = tich * (1.0-yy);
            }
        }
        tich = 1 - tich;
        cout << fixed << setprecision(6) << tich << endl;

    }
}