Solve in Python:
You are given an array $a$ consisting of $n$ integers. You have to find the length of the smallest (shortest) prefix of elements you need to erase from $a$ to make it a good array. Recall that the prefix of the array $a=[a_1, a_2, \dots, a_n]$ is a subarray consisting several first elements: the prefix of the array $a$ of length $k$ is the array $[a_1, a_2, \dots, a_k]$ ($0 \le k \le n$).

The array $b$ of length $m$ is called good, if you can obtain a non-decreasing array $c$ ($c_1 \le c_2 \le \dots \le c_{m}$) from it, repeating the following operation $m$ times (initially, $c$ is empty):  select either the first or the last element of $b$, remove it from $b$, and append it to the end of the array $c$. 

For example, if we do $4$ operations: take $b_1$, then $b_{m}$, then $b_{m-1}$ and at last $b_2$, then $b$ becomes $[b_3, b_4, \dots, b_{m-3}]$ and $c =[b_1, b_{m}, b_{m-1}, b_2]$.

Consider the following example: $b = [1, 2, 3, 4, 4, 2, 1]$. This array is good because we can obtain non-decreasing array $c$ from it by the following sequence of operations:  take the first element of $b$, so $b = [2, 3, 4, 4, 2, 1]$, $c = [1]$;  take the last element of $b$, so $b = [2, 3, 4, 4, 2]$, $c = [1, 1]$;  take the last element of $b$, so $b = [2, 3, 4, 4]$, $c = [1, 1, 2]$;  take the first element of $b$, so $b = [3, 4, 4]$, $c = [1, 1, 2, 2]$;  take the first element of $b$, so $b = [4, 4]$, $c = [1, 1, 2, 2, 3]$;  take the last element of $b$, so $b = [4]$, $c = [1, 1, 2, 2, 3, 4]$;  take the only element of $b$, so $b = []$, $c = [1, 1, 2, 2, 3, 4, 4]$ — $c$ is non-decreasing. 

Note that the array consisting of one element is good.

Print the length of the shortest prefix of $a$ to delete (erase), to make $a$ to be a good array. Note that the required length can be $0$.

You have to answer $t$ independent test cases.


-----Input-----

The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.

The first line of the test case contains...
def solve():
    n = int(input())
    a = list(map(int, input().split()))
    prev = a[-1]
    cnt = 1
    status = 0
    for i in range(n - 2, -1, -1):
        v = a[i]
        if status == 0:
            if v < prev:
                status = 1
            cnt += 1
            prev = v
            continue
        elif status == 1:
            if v > prev:
                break
            cnt += 1
            prev = v
            continue
    print(n - cnt)

t = int(input())
for _ in range(t):
    solve()