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.*;
 
public class A {
 
	public static void main(String[] args)throws IOException 
	{
		BufferedReader ob = new BufferedReader(new InputStreamReader(System.in));
		StringBuffer sb = new StringBuffer();
		
		int test=Integer.parseInt(ob.readLine());
		while(test-->0)
		{
			int n=Integer.parseInt(ob.readLine());
			String s[]=ob.readLine().split(" ");
			
			if(n==1)
			{
               sb.append("0\n");
               continue;
            }
			
            long arr[]=new long[n];
            long max=Integer.MIN_VALUE;
            ArrayList<Long> d = new ArrayList<Long>();
            
            for(int i=0;i<n;i++)
            {
                arr[i]=Long.parseLong(s[i]);
                d.add(max-arr[i]);
                
                if(arr[i]>max)
                  max=arr[i];
            }
            if(Collections.max(d)<=0)
            {
                sb.append("0\n");
                continue;
            }
       
            long temp=Collections.max(d);
            int ans=highestPowerof2(temp)+1;
            
            sb.append(ans+"\n");
		}
            
		System.out.println(sb);
	}
	
	static int highestPowerof2(long n) 
	{    
    	int p = (int)(Math.log(n)/Math.log(2)); 
	    return p;
	} 
	static long fastPow(long n, long m) {
		long res = 1L;
 
		while (m > 0) {
			if ((m & 1) != 0) {
				res *= n;
				m--;
			} else {
				n *= n;
				m /= 2;
			}
		}
		return res;
	}
}