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}, …, 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, …, 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} ≤ a_{2} ≤ … ≤ a_{n}.

You have to answer t independent test cases.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^{4}) — the number of test cases.

The first line of each test case contains single integer n (1 ≤ n ≤ 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}, …, a_{n} (-10^{9} ≤ a_{i} ≤ 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].
def get_power(n):
    if n == 1: 
        return 0
    
    return 1 + get_power(int(n / 2))


t = int(input())
res = []

for i in range(0, t):
    input()
    arr = [int(j) for j in input().strip().split()]

    if len(arr) <= 1:
        res.append(0)
        continue

    min_element = arr[len(arr) - 1]
    max_diff = arr[len(arr) - 2] - arr[len(arr) - 1]

    for j in reversed(range(0, len(arr) - 1)):
        cur = arr[j]
        max_diff = max(max_diff, cur - min_element)
        min_element = min(cur, min_element)
    
    if max_diff <= 0:
        res.append(0)
    else:
        result = get_power(max_diff)
        if 2**(result + 1) >= max_diff:
            res.append(1 + result)
        else: 
            res.append(2 + result)

for e in res:
    print(e)