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 <algorithm>
#include <array>
#include <cassert>
#include <cmath>
#include <cstring>
#include <deque>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <memory>
#include <optional>
#include <queue>
#include <set>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

using namespace std;

using big = long long;

int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  int cas;
  cin >> cas;
  while (cas--) {
    int n, m;
    cin >> n >> m;
    vector<int> a(n), sorted(n + 1);
    vector<pair<int, double>> b(m);
    for (int &v : a) {
      cin >> v;
    }
    for (auto &v : b) {
      cin >> v.first >> v.second;
      --v.first;
    }
    reverse(b.begin(), b.end());
    sorted[n] = true;
    int mn = n;
    for (int i = n - 1; i >= 0; --i) {
      sorted[i] = sorted[i + 1];
      if (a[i] != i + 1) {
        sorted[i] = false;
      }
      if (sorted[i]) {
        mn = i;
      }
    }
    // for (int i = 0; i <= n; ++i) {
    //   cerr << sorted[i] << " ";
    // }
    // cerr << endl;
    double ans = 1;
    for (const auto &op : b) {
      int i = op.first;
      double p = op.second;
      if (i + 1 >= mn) {
        ans *= 1 - p;
      }
    }
    ans = 1 - ans;
    if (sorted[0]) {
      ans = 1;
    }
    cout << fixed << setprecision(8) << ans << "\n";
  }
}