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].
// Akash Yadav(aky91)

import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;

public class Solution{

	static Scanner scn = new Scanner(System.in);
	static boolean multipleTC = true;

	public static void main (String[] args) {
		int T = (multipleTC)?scn.nextInt():1;
		for(int t = 1; t <= T; t++) solve(t);
	}

	static void solve(int tcno){
		int N = ni();
		int[] arr = rai(N);
		
		int deepest = 0;
		int high = arr[0];
		int low = arr[0];
		for(int i = 1; i < N; i++){
		    int n = arr[i];
		    if(n > high){
		        high = n;
		        low = n;
		    }
		    
		    if(n < low){
		        low = n;
		    }
		    
		    int diff = high - low;
		    if(diff > deepest) deepest = diff;
		}
		
		if(deepest == 0){
		    pln(0); 
		    return;
		}
		
		pln((int)(Math.log(deepest) / Math.log(2))+1);
	}

    //SUPPORTING FUNCTIONS
	static void p(Object o){System.out.print(o + " ");}
	static void pln(Object o){System.out.println("" + o);}
	static void pln(){System.out.println();}
	static int ni(){return scn.nextInt();}
	static double nd(){return scn.nextDouble();}
	static long nl(){return scn.nextLong();}
	static String ns(){return scn.next();}
	static String nline(){return scn.nextLine();}
	static int[] rai(int n){int[] arr = new int[n]; for(int i = 0; i < n; i++) arr[i] = ni(); return arr;}
	static long[] ral(int n){long[] arr = new long[n]; for(int i = 0; i < n; i++) arr[i] = nl(); return arr;}
	static int[][] raai(int n, int m){int[][] arr = new int[n][m]; for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) arr[i][j] = ni(); return arr;}
}