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 no_diamonds(n):
    return n

def problem1():
    print(no_diamonds(int(input())))

def sort_by_adjsum(a,b):
    l = len(a) 
    if l == 1:
        return a + b
    else:
        return sort_by_adjsum(a[l//2:],b[:(l-1)//2 + 1]) + sort_by_adjsum(a[:l//2],b[(l-1)//2 + 1:])

def problem2():
    l = int(input())
    a = list(map(int,input().split()))
    a.sort()
    b = sort_by_adjsum(a[:l//2],a[(l-1)//2+1:])
    if l%2 == 1:
        b = [a[l//2]] + b
    print(*b)

def powadd(a):
    s = 0
    t = 0
    u = 1
    b = a[0]
    for i in range(1,len(a)):
        if b <= a[i]:
            b = a[i]
        elif b > a[i] and b <= a[i] + s:
            continue
        else:
            while b > a[i] + s:
                s += u
                u *= 2
                t += 1
    return t

def problem3():
    l = int(input())
    a = list(map(int,input().split()))
    print(powadd(a))

cases = int(input())
for _ in range(cases):
    # problem1()
    # problem2()
    problem3()