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:
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
#define I_love_andrey_sandimirov cout.precision(20); ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define pb push_back
#define forn(i, n) for(ll (i) = 0; (i) < (n); ++(i))
#define fornm(i, m, n) for(ll (i) = (m); (i) < (n); ++(i))
#define rfornm(i, m, n) for(ll (i) = (m); (i) >= (n); --(i))
#define readvec(vector, n) {ll temp_vec_val;forn(inc, n){cin >> temp_vec_val;vector.push_back(temp_vec_val);}}
#define printvec(vector, delimeter) {ll length_of_vector=vector.size();\
                                    forn(elementvec, length_of_vector){cout << vector[elementvec] << delimeter;}}
#define all(vector) (vector).begin(), (vector).end()
#define rall(vector) (vector).rbegin(), (vector).rend()
#define endl "\n"
#define sz(something) (ll)something.size()
const ld EPS = 1e-8;
const ll MOD = 1000000007;
const ll INF = 1e18;
const ld pi = 2.0*acos(0.0);
ll n, m, k, p, q, t, sum = 0, cnt = 0;
ll mx = -INF;
ll mn = INF;
bool flag = 0;
vll v;
string s = "", s1, s2;
char c;

void solve()
{
    cin >> t;
    while(t--)
    {
        cin >> n >> m;
        v.clear();
        readvec(v,n);
        vector<pair<ll,ld>> vv;
        forn(i,m)
        {
            ld q;
            cin >> p >> q;
            vv.pb({p,q});
        }
        sort(rall(vv));
        p = n-1;
        while(p>=0&&v[p]==p+1)--p;
        if(p==-1)
        {
            cout << 1 << endl;
            continue;
        }
        ld cn=1.;
        forn(i,m)
        {
            if(vv[i].first<=p) break;
            cn*=1.-vv[i].second;
        }
        cout << fixed << setprecision(10) << 1.0-cn << endl;
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("H:\\repos\\projects\\contests_vsc\\in.txt", "r", stdin);
    freopen("H:\\repos\\projects\\contests_vsc\\out.txt", "w", stdout);
#endif
    I_love_andrey_sandimirov
    solve();
    return 0;
}