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:
import logging
import sys
from inspect import currentframe

sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline

logging.basicConfig(level=logging.DEBUG)


def dbg(*args):
    id2names = {id(v): k for k, v in currentframe().f_back.f_locals.items()}
    logging.debug(
        ", ".join(id2names.get(id(arg), "???") + " = " + repr(arg) for arg in args)
    )


def solve():
    n, m = map(int, input().split())
    a = list(map(int, input().split()))
    exps = [tuple(map(float, input().split())) for _ in range(m)]
    exps.sort()
    limit_ri = 0  # このindex以下の要素を必ずソートしないといけない
    for i in range(n - 1, -1, -1):
        if a[i] != i + 1:
            limit_ri = i
            break
    else:  # 最初からソートされている場合
        return 1
        # for i in range(n):
    #     s += a[i]
    #     if s == (i + 1) * (i + 2) // 2:
    #         if unsorted:
    #             limit_ri = i
    #     else:
    #         unsorted = True
    # if not unsorted:
    #     return 1
    # dbg(limit_ri)

    unsort_prob = 1
    for ri, pi in exps:
        if ri > limit_ri:
            unsort_prob *= 1 - pi
    return 1 - unsort_prob


def main():

    t = int(input())
    for tt in range(t):
        print(solve())


if __name__ == "__main__":
    main()