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).

<image>

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 pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define REP(i,n) for(int _n=n, i=0;i<_n;++i)
#define FOR(i,a,b) for(int64_t i=(a),_b=(b);i<=_b;++i)
#define FORD(i,a,b) for(int64_t i=(a),_b=(b);i>=_b;--i)
using ull = uint64_t;
using ll = int64_t;
using PII = pair<int, int>;
using VI = vector<int>;

string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string) s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B> string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; }
template <typename A> string to_string(A v) {
bool first = true; string res = "{";
for (const auto &x : v) { if (!first) { res += ", "; } first = false; res += to_string(x); }
res += "}"; return res; }
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail> void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H); debug_out(T...); }
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)

const ll md = 1000000007;

using ld = long double;

void solve() {
	int n, m;
	cin >> n >> m;
	VI V(n);
	REP(i, n) cin >> V[i];

	int idx = n + 1;
	FORD(i, n, 1) {
		if (V[i - 1] != i) {
			idx = i;
			break;
		}
	}
	ld pp = 1.0;
	while (m--) {
		int k;
		ld p;
		cin >> k >> p;
		if (k >= idx) {
			p = 1.0 - p;
			pp *= p;
		}
	}
	if (idx == n + 1) {
		cout << "1.0\n";
		return;
	}
	std::cout << std::fixed;
    std::cout << std::setprecision(9);
	cout << 1.0 - pp << "\n";
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    int t;
    cin >> t;
    while (t--) solve();
}