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;
#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c> {i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
  ~debug() { cerr << endl; }
  eni( != ) cerr << boolalpha << i; ris;
}
eni( == ) ris << range(begin(i), end(i));
}
sim, class b dor(pair < b, c > d) {
  ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
  *this << "[";
  for (auto it = d.b; it != d.e; ++it)
    *this << ", " + 2 * (it == d.b) << *it;
  ris << "]";
}
#else
  sim dor(const c&) { ris; }
#endif
};
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "
///////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////
#define int long long
#define double long double

void solve() {
  int n, m;
  cin >> n >> m;
  vector<int>a(n);
  int suff = n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] != i + 1) {
      suff = (i + 1);
      break;
    }
    if (i == 0) {
      suff = 0;
    }
  }
  vector<pair<int, double>>vals(m);
  for (int i = 0; i < m; i++) {
    cin >> vals[i].first;
    cin >> vals[i].second;
  }
  sort(vals.rbegin(), vals.rend());
  if (suff == 0) {
    cout << fixed << setprecision(6) << 1.000000 << endl;
    return;
  }

  double prob = 0.000000;
  double nxt = 1.000000;
  for (int i = 0; i < m; i++) {
    if (vals[i].first < suff) {
      break;
    }
    if (i == 0) {
      prob = vals[i].second;
      nxt = (1.000000 - prob);
    }
    else {
      prob += (nxt * vals[i].second);
      nxt *= (1.000000 - vals[i].second);
    }

  }

  cout << fixed << setprecision(6) << prob << endl;




}


int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int t;
  cin >> t;
  while (t--)
    solve();

}