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].
from math import log
for _ in range(int(input())):
    n = int(input())
    a = [int(x) for x in input().split()]
    #b = [x ** 2 for x in range(100000)]
    mx = 0
    #mn = 100000000
    #ans = 0
    maxdif = 0
    i = 1
    start = a[0]
    while i < n:
        start = a[i - 1]
        if a[i - 1] > a[i]:
            #print(f'from {a[i - 1]}')
            while i < n and start > a[i]:
                maxdif = max(maxdif, start - a[i])
                i += 1
            #print(f'to {a[i - 1]}, maxdif = {maxdif}')
        else:
            i += 1


    print(len(bin(maxdif)) - 2 if maxdif > 0 else 0)


            #maxdif = max(a[i - 1] - a[i], maxdif)
    #for i in range(10000000):
    #    if 2 ** (i - 1) >= maxdif:
    #        print(i)
    #        break
    #x = 0
    #i = 0
    #while x < maxdif:
    #    x += 2 ** i
    #    i += 1
    #print(i)
            #for j in range(len(b)):
                #if a[i - 1] <= a[i] + 2 ** b[j]:
                    #a[i] += 2 ** b[j]
                    #mx = max(mx, j)
                    #mn = min(mn, j)
                    #b[j] = 0
                    #break
    #print(max(0, mx - mn))
    #print(mx)
    #print(a)
