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.*;
import java.io.*;
public class Addition {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        for (int i1 = 0; i1 < t; i1 ++) {
            int n = sc.nextInt();
            int[] arr = new int[n];
            for (int i2 = 0; i2 < n; i2 ++){
                arr[i2] = sc.nextInt();
            }
            int T = 0;
            for (int i2 = 1; i2 < n; i2 ++){
                int diff = arr[i2] - arr[i2 - 1];
                if (diff < 0){
                    int seconds = 0;
                    for (int x = 0; x < 60; x ++){
                        if (Math.pow(2, x) > -diff){
                            seconds = x;
                            break;
                        }
                    }
                    if (seconds > T){
                        T = seconds;
                    }
                    arr[i2] = arr[i2 - 1];
                }
            }
            pw.println(T);
        }
        pw.close();
    }
}

