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.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t;
        t = in.nextInt();
        int i = 1;
            long ans = 1;
            TreeMap<Long, Integer> x = new TreeMap<>();
            while (ans < Long.MAX_VALUE ) {
                ans = (1l<<i) - 1;
                x.put(ans, i);
                i++;
            }
        while((t--) >0 ){
            int n;
            n = in.nextInt();
            int[] arr = new int[n];
            for( i = 0;i<n;i++){
                arr[i] = in.nextInt();
            }
            
            long max = 0;
            for(int j =1; j< n;j++){
                if(arr[j-1] > arr[j]){
                    int diff = arr[j-1] - arr[j];
                    int val  = x.ceilingEntry((long)diff).getValue();
                    max = Math.max(max,val);
                    //   System.out.println(arr[j]);
                    arr[j] += diff;
                    // System.out.println("val " +  val  +   "   " + arr[j]);
                }
            }
            // for(int j = 0; j<n; j++){
            //     System.out.println(arr[j]);
            // }
            // System.out.println("-------");
            System.out.println(max);
        }

    }
}
