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.*;
public class Main {

  public static void main(String[] args) 
  {
  
	    Scanner ab=new Scanner(System.in);
	    int q=ab.nextInt();
	    long pow[]=new long[41];
	    for(int i=0;i<=40;i++)
	    {pow[i]=(long)Math.pow(2,i);}
	    //System.out.println(Arrays.toString(pow));
	    while(q-->0)
	    {
	     
	     int x=ab.nextInt();
	     int mx=0;
	     long arr[]=new long[x];
	     long dif[]=new long[x-1];
	     arr[0]=ab.nextLong();
	     for(int i=1;i<x;i++)
	     {
	     arr[i]=ab.nextLong();
	     
	     
	     dif[i-1]=arr[i]-arr[i-1];
	     if(dif[i-1]<0)
	     {
	       
	       long ele=dif[i-1]*-1;
	       for(int s=0;s<41;s++)
	       {
	         
	         if(pow[s]>ele)
	         {
	         arr[i]=arr[i-1];
	         mx=Math.max(s,mx);
	         break;
	         }
 	         
           }
	       
	       
         }
	     }
	     //if(mx>0)mx++;
	     System.out.println(mx);
	      
        }
	}
}

