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>

#define ALL(x) x.begin(), x.end()
#define pb push_back
#define db1(x) cout << #x << " = " << x << endl
#define db2(x, y) cout << #x << " = " << x << ", " << #y << " = " << y << endl
using namespace std;
void localFileInput() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
}
typedef long long ll;
typedef pair<int, int> pii;
const ll mod = 1e9+7;
const int N = 1e5+10;
int n, m, a[N];
struct Node
{
    int r;
    double p;
    void input() {
        scanf("%d%lf", &r, &p);
    }
}b[N];
int r[N];
bool cmp(Node a , Node b) {
    return a.r < b.r;
}
void sol() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++ i) scanf("%d", &a[i]);
    for (int i = 1; i <= m; ++ i) b[i].input();
    sort(b + 1, b + 1 + m, cmp);
    for (int i = 1; i <= m; ++ i) r[i] = b[i].r;
    int id = 0;
    for (int i = n; i >= 1; -- i) if(a[i] != i) {
        id = i;
        break;
    }
    if(id == 0) {
        puts("1.000000");
        return ;
    }
    int st = lower_bound(r+1, r+1+m, id) - r;

    double ans = 1.0;
    for (int i = st; i <= m; ++ i) 
        ans = ans * (1 - b[i].p);
    printf("%.6f\n", 1-ans);
}
int main() {
    localFileInput();
    int _; scanf("%d", &_); while(_ --) sol();
    return 0;
}