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].
//package cc;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
import java.util.*;
import java.util.Collections;
import java.util.Scanner;
import java.util.*;
public class cd {
  static int bits(long n)
  {int bts=0;
	  while(n>0)
	  {n=n/2;
	  ++bts;
		  
	  }
	  return bts;
  }

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s=new Scanner(System.in);
        int t=s.nextInt();
       
        while(t-->0)
        {
        	int n;
        	
        	n=s.nextInt();
        	long max=0;
        	long ar[]=new long[n];
        	for(int i=0;i<n;++i)
        		ar[i]=s.nextLong();
        	
        	for(int i=0;i<n;++i)
        	{  if(i==n-1)
        		continue;
        		if(ar[i+1]>=ar[i])
        			continue;
        		long k=ar[i];
        		long m=0;
        		++i;
        		while(i<n&&ar[i]<k)
        		{
        			if(k-ar[i]>m)
        				m=k-ar[i];
        			++i;
        			
        		}
        		--i;
        		
        		
        		if(m>max)
        			max=m;
        			
        		
        	}
        	     System.out.println(bits(max));	
        }
        

	}

}
