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:
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("O2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
////
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
#include <bits/stdc++.h>

#define all(x) (x).begin(), (x).end()
#define allp(x) (x)->begin(), (x)->end()

#define pb push_back


using namespace std;
void dout() { cerr << endl; }
//typedef long long ll;
template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
    cerr << H << ' ';
    dout(T...);
}

using ll = long long;
//#ifdef __int128
//using hll = __int128;
//#endif
using pii = pair<ll, ll>;
using ld = long double;

template <typename T>
void do_uniq(vector<T> &vec){
    sort(all(vec));
    vec.resize(unique(all(vec)) - vec.begin());
}

#ifndef ONLINE_JUDGE

clock_t timestamp_start = clock();
void time_calc()
{
    cerr << (ld)(clock() - timestamp_start) / CLOCKS_PER_SEC << "\n";
}
#endif


#ifdef _getchar_nolock
#else
#    define _getchar_nolock getchar_unlocked
#endif

#define integer int

integer mod = 1e9 + 7;

integer ml(integer a, integer b) {
    return (a * 1ll * b) % mod;
}

integer add(integer a, integer b) {
    return (a + b) % mod;
}

integer sub(integer a, integer b) {
    return add(a, mod - b);
}

integer sq(integer a) {
    return ml(a, a);
}

integer b_p(integer b, integer p) {
    if (p == 0) return 1;
    if (p & 1) return ml(b, b_p(b, p - 1));
    return sq(b_p(b, p >> 1));
}

integer bback(integer e) {
    return b_p(e, mod - 2);
}

#define solvsh
#define multi

#ifdef solvsh

const int MAXN = 1e3;

void solve() {

    int n, m;
    cin >> n >> m;

    ld answr = 1;

    vector<int> v;

    for (int i = 0; i < n; ++i) {

        int curr;
        cin >> curr;

        --curr;

        v.push_back(curr);
    }
    bool ok = true;
    for (int i = 0; i < n; ++ i) {
        if (v[i] != i) ok = false;
    }
    if (ok) answr = 0;

    int maxxx = n;
    while (maxxx > 0 && v[maxxx - 1] == maxxx - 1) --maxxx;


    for (int i = 0; i < m; ++i) {

        int r;
        ld pi;
        cin >> r >> pi;

        if (r < maxxx) continue;

        answr *= 1 - pi;
    }

    cout << 1 - answr << "\n";

}

#else

#endif

void multisolve() {
//    precalc();

    int t;
    cin >> t;

    int i = 1;
    while (t--) {
        solve();
        i++;
    }
}

#ifndef ONLINE_JUDGE

void gen() {

    int n = 2000;
    cout << n << "\n";
    for (int i = 0; i < n; ++i) {

        for (int j = 0; j < n; ++j) {
            if (i == 0 && j == 0) {
                cout << "S";
                continue;
            }
            if (i == n - 1 && j == n  - 1) {
                cout << "G";
                continue;
            }

            cout << char('a' + (rand() % 26)) << " ";
        }
        cout << "\n";
    }
}


#endif

#define int int
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    cout << fixed << setprecision(20);

#ifndef ONLINE_JUDGE
#define NAME "../input"
    freopen(NAME".in", "r", stdin);
//    freopen("../output.out", "w", stdout);
#else
#endif
#ifdef multi
    //            gen();
        multisolve();
#else

    solve();
//    gen();
#endif
//     time_calc();
}