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].
"""

2^x-2 <= a <= 2^x-1

x-2 <= log2(a) <= x-1

x-1 <= log2(a)+1 <= x

x <= log2(a)+2 <= x+1


1, 2, 4, 8, 16, 32


1 0 -2 -5 -9 -16 -24 -39
b       x = floor(log(b, 2))+1     plus
1       1                           1
2~3     2                           2 / 4
4~7     3                           4 / 8
8~15    4
16~31   5

2^(n-1) ~ 2^n - 1    n번
2^(n-1)<= b < 2^n
n-1 <= log2(b) < n
n <= log2(b)+1 < n+1    
-11? +16
-8? +8
-9? +16
-7? +8

log2(11) = 3.4
"""
from math import log2
import sys
input = sys.stdin.buffer.readline
for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    res = 0
    prev = a[0]
    for i in range(1, n):
        b = a[i]-prev
        if b >= 0:
            prev = a[i]
        else:
            x = int(log2(-b) + 1)
            res = max(res, x)
    print(res)