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 random
from math import log2, ceil, log
input = sys.stdin.readline

def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp
def gcd(a,b):
    if a > 0:
        return gcd(b%a, a)
    return b
def primefs(n):
    ## if n == 1
    if n == 1:
        return 1
    ## calculating primes
    primes = {}
    while(n%2 == 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    return primes
    ## prime factoriazation of n is stored in dictionary primes

## DISJOINT SET UNINON FUNCTION
def swap(a, b):
    temp = a
    a = b
    b = temp
    return a,b
# find function
def find(x, link):
    while(x != link[x]):
        x = link[x]
    return x
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, size, link):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    return prime

## taking integer array input
def int_array():
    return list(map(int, input().split()))

############ ---------------- TEMPLATE ENDS HERE ---------------- ############


for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    if n == 1:
        print(0)
        continue;
    maxi = a[0]; ans_max = 0
    for i in range(1, n):
        if a[i] < maxi:
            ans_max = max(ans_max, maxi - a[i]);
        maxi = max(maxi, a[i])
    count = 0;
    while(ans_max > 0):
        ans_max = ans_max - 2**count;
        count += 1;
    print(count)
