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].
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###################################The_Apurv_Rathore#####################################################
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import sys,os,io
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop


import math 



def ncr(n, r, p):  #using fermat's little theorem
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,
                      p - 2, p)) % p

  
def primeFactors(n): 
    l = []
    while n % 2 == 0: 
        l.append(2)
        n = n / 2
          
    for i in range(3,int(math.sqrt(n))+1,2): 
          

        while n % i== 0: 
            l.append(int(i))
            n = n / i 
    if n > 2: 
        l.append(n)
    return list(set(l))
    
def power(x, y, p) : 
	res = 1
	x = x % p 
	if (x == 0) : 
		return 0
	while (y > 0) : 
		if ((y & 1) == 1) : 
			res = (res * x) % p 
		y = y >> 1	 # y = y/2 
		x = (x * x) % p 		
	return res 


def si():
    return input()

# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

def prefix_sum(arr):
    r = [0] * (len(arr)+1)
    for i, el in enumerate(arr):
        r[i+1] = r[i] + el
    return r

def divideCeil(n,x):
    if (n%x==0):
        return n//x
    return n//x+1

def ii():
    return int(input())

def li():
    return list(map(int,input().split()))

def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')


if(os.path.exists('input.txt')):
    sys.stdin = open("input.txt","r")
    sys.stdout = open("output.txt","w")
t = 1
t = int(input())
for _ in range(t):
    n = ii()
    l = li()
    res = 0
    m = 0
    mm = l[0]
    for i in range(n-1):
        mm = max(mm,l[i])
        if (mm>l[i+1]):
            # print("res,m",res,m)
            if (m<mm-l[i+1]):
                res+=mm-l[i+1]-m
                m = mm-l[i+1]
            else:
                continue
        else:
            continue
    x = 0
    if (res==0):
        print(0)
        continue
    # print(l)
    for i in range(31):
        res-=2**x
        x+=1
        # print("res",res)
        if (res<=0):
            break
    print(x)

    

