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 PYTHON3):

Solution:
from queue import LifoQueue


def print_sol(c, sol):
    print('Case #{}: {}'.format(c, sol))


def rd_int():
    return int(input())


def rd_str():
    return input()


def rd_str_lst():
    return input().split(' ')


def sp_str_lst():
    return list(input())


def rd_int_lst():
    return [int(i) for i in rd_str_lst()]


def rd_float_lst():
    return [float(i) for i in rd_str_lst()]


def sp_int_lst():
    return [int(i) for i in sp_str_lst()]


def dfs(g, i, v):
    """
    recursive post order dfs
    Args:
        g: array of graph nodes
        i: node to traverse
        v: visited array

    Returns: a post order dfs of graph g
    """
    v[i] = True
    for j in g[i]:
        if not v[j]:
            dfs(g, j, v)

    print(i)


def dfs_iter(g, i, n):
    v = [False] * n
    stack = LifoQueue()
    stack.put(i)
    while stack.not_empty:
        u = stack.get()
        v[u] = True
        print(u)
        for i in g[u]:
            if not v[i]:
                stack.put(i)


def happiness(g, i, v, p, h, f):
    au = p[i]
    gv = 0

    v[i] = True
    for j in g[i]:
        if not v[j]:
            av, gvi, f = happiness(g, j, v, p, h, f)
            au += av
            gv += gvi

    if not (au + h[i]) % 2 == 0:
        f = False

    gu = (au + h[i]) // 2

    if not 0 <= gu <= au:
        f = False

    if not gu >= gv:
        f = False

    return au, gu, f


def main():
    t = rd_int()

    for _ in range(1, t + 1):
        n, m = rd_int_lst()
        a = rd_int_lst()
        sp = [rd_float_lst() for i in range(m)]

        x = n
        while a[x - 1] == x and x > 0:
            x -= 1

        if x == 0:
            print(1)
            continue

        p = 1
        s = 0
        for i in range(m):
            if sp[i][0] >= x:
                s += p * sp[i][1]
                p *= (1 - sp[i][1])
        print(s if p < 1 else 0)


if __name__ == '__main__':
    main()