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.io.*;
import java.util.*;

final public class A {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int tst = Integer.parseInt(br.readLine());
        while(tst-->0){
            int n = Integer.parseInt(br.readLine());
            StringTokenizer st = new StringTokenizer(br.readLine());
            long[] arr = new long[n];
            for(int i = 0; i<n; i++){
                arr[i] = Integer.parseInt(st.nextToken());
            }
            long max_till_now = arr[0], max_diff = 0, t = 0;
            for(int i = 0; i+1<n; i++){
                long diff = max_till_now - arr[i+1];
                if(diff>0){
                    if(diff>max_diff){
                        long p = (long)Math.pow(2, t)+max_diff;
                        t++;
                        while(diff>p){
                            p+=(long)Math.pow(2, t);
                            t++;
                        }
                        max_diff = p;
                    }
                    arr[i+1] = max_till_now;
                }
                else max_till_now = arr[i+1];
            }
            sb.append(t).append('\n');
        }
        System.out.println(sb);
    }
}