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 java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Solution1338A {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int t = scanner.nextInt();

        while (t-- > 0)
        {
            int n = scanner.nextInt();

            List<Long> list = new ArrayList<>();

            for (int i = 0; i < n; ++i)
            {
                list.add(scanner.nextLong());
            }

            solve(list);
        }
    }

    public static void solve(List<Long> list)
    {
        long maxTime = 0;

        for (int i = 1; i < list.size(); ++i)
        {
            if (list.get(i) < list.get(i-1))
            {
                maxTime = Math.max(maxTime, list.get(i-1) - list.get(i));
                list.set(i, list.get(i-1));
            }
        }

        int ans = (int) (Math.ceil(log2(maxTime + 1)));
        System.out.println(ans);
    }


    private static double log2(long x)
    {
        return (Math.log(x) / Math.log(2));
    }


}
