You have an array $a$ of length $n$. For every positive integer $x$ you are going to perform the following operation during the $x$-th second:
  Select some distinct indices $i_{1}, i_{2}, \ldots, i_{k}$ which are between $1$ and $n$ inclusive, and add $2^{x-1}$ to each corresponding position of $a$. Formally, $a_{i_{j}} := a_{i_{j}} + 2^{x-1}$ for $j = 1, 2, \ldots, k$. Note that you are allowed to not select any indices at all. 
You have to make $a$ nondecreasing as fast as possible. Find the smallest number $T$ such that you can make the array nondecreasing after at most $T$ seconds.
Array $a$ is nondecreasing if and only if $a_{1} \le a_{2} \le \ldots \le a_{n}$.
You have to answer $t$ independent test cases.
-----Input-----
The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases.
The first line of each test case contains single integer $n$ ($1 \le n \le 10^{5}$) — the length of array $a$. It is guaranteed that the sum of values of $n$ over all test cases in the input does not exceed $10^{5}$.
The second line of each test case contains $n$ integers $a_{1}, a_{2}, \ldots, a_{n}$ ($-10^{9} \le a_{i} \le 10^{9}$).
-----Output-----
For each test case, print the minimum number of seconds in which you can make $a$ nondecreasing.
-----Example-----
Input
3
4
1 7 6 5
5
1 2 3 4 5
2
0 -4
Output
2
0
3
-----Note-----
In the first test case, if you select indices $3, 4$ at the $1$-st second and $4$ at the $2$-nd second, then $a$ will become $[1, 7, 7, 8]$. There are some other possible ways to make $a$ nondecreasing in $2$ seconds, but you can't do it faster.
In the second test case, $a$ is already nondecreasing, so answer is $0$.
In the third test case, if you do nothing at first $2$ seconds and select index $2$ at the $3$-rd second, $a$ will become $[0, 0]$.
for _ in range(int(input())):
    n = int(input())
    l1 = list(map(int, input().split()))
    answer = 0
    max_so_far = l1[0]
    for i in range(1, n):
        if l1[i] >= max_so_far:
            max_so_far = l1[i]
        else:
            answer = max(answer, (max_so_far - l1[i]).bit_length())
        
        
    print(answer)
for _ in range(int(input())):
    n = int(input())
    l1 = list(map(int, input().split()))
    answer = 0
    max_so_far = l1[0]
    for i in range(1, n):
        if l1[i] >= max_so_far:
            max_so_far = l1[i]
        else:
            answer = max(answer, (max_so_far - l1[i]).bit_length())
        
        
    print(answer)