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].
import sys
import math
input = sys.stdin.buffer.readline
def I(): return(list(map(int,input().split())))
def sieve(n):
	a=[1]*n
	for i in range(2,n):
	    if a[i]:
	        for j in range(i*i,n,i):
	            a[j]=0
	return a

for __ in range(int(input())):
	n=int(input())
	arr=I()
	if n==1:print(0)
	else:
		maxdiff=0
		mins=[0]*n
		m=arr[-1]
		usedic={}
		mins[-1]=m
		for i in reversed(range(n-1)):
			mins[i]=min(mins[i+1],arr[i])
		i=0
		while(i<n-1):
			if arr[i]<=arr[i+1]:
				i+=1
				continue
			else:
				maxdiff=max(arr[i]-mins[i],maxdiff)
				usedic[maxdiff]=usedic.get(arr[i]-mins[i],0)+1
				i+=1
		if maxdiff==0:print(0)
		else:
			# print(maxdiff,usedic)

			ans=math.log(maxdiff,2)
			if ans==math.ceil(ans):
				ans=math.ceil(ans)+1
			else:
				ans=math.ceil(ans)
			# if usedic[maxdiff]>1:ans+=usedic[maxdiff]-1
			print(ans)

		# if i==n-1:print(0)
		# else:
		# 	j=i
		# 	i+=1
		# 	m=arr[i]
		# 	while(i<n):
		# 		m=min(m,arr[i])
		# 		i+=1
		# 	print(math.ceil(math.log(arr[j]-m,2))+1)
		






