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.ArrayList;
import java.util.Collections;
import java.util.Scanner;
public class C633C
{
    static long power[];
    public static void main(String[] args)
    {
        power=new long[64];
        power[0]=1;
        for(int x=1;x<64;x++)
        {
            power[x]=power[x-1]*2l;
        }
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0)
        {
            int n=sc.nextInt();
            long[] ar =new long[n];
            for(int x=0;x<n;x++)
            {
                ar[x]=sc.nextLong();
            }
            solve(ar,n);
        }
    }

    public static void solve(long[] ar, int n)
    {
        long ans=0;
        for(int x=1;x<n;x++)
        {
            if(ar[x]<ar[x-1])
            {
                long c=0;
                long d=ar[x-1]-ar[x];
                while(d>0)
                {
                    d/=2;
                    c++;
                }
                ar[x]=ar[x-1];
                ans=Math.max(ans,c);
            }
        }
        System.out.println(ans);
    }
}