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].
t = int(input())

for p in range(0,t):

    n = int(input())
    l = list(map(int,input().split()))
    mx = l[0]
    res = 0
    for i in range(1,n):
        res = max(res,mx-l[i])
        mx = max(mx,l[i])
    cnt = 0
    while res>0:
        cnt+=1
        res//=2
    print(cnt)









    # diff = []
    # for i in range(1,n):
    #     if(a[i] - a[i-1]<0):
    #         diff.append(-1* (a[i]-a[i-1]))
    # if(len(diff) == 0):
    #     print(0)
    # else:
    #     diff = sorted(diff)[::-1]
    #     print(diff)
    #     p = 1
    #     while(2**(p-1) < diff[len(diff)-1]):
    #         p+=1
    #
    #     x = 1
    #     while(2**(x-1) < diff[0]):
    #         x+=1
    #
    #     print(p,x)
    #     print(2**30,2**29)
    #
    #
    #     if(len(diff) <= x-p +1):
    #         print(x)
    #     else:
    #         print(x + (len(diff) - (x-p+1)))




        # largestcuont = diff.count(diff[0])
        # x = x + (largestcuont -1)
        #
        # i = 1
        # while(diff[i] == diff[0]):
        #     i+=1
        #
        # p = 1
        # while(2**(p-1) < diff[len(diff)-1]):
        #     p+=1
        #
        # if(len(diff[i:]) > x):







