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.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;


public  class FibinacciSeries 
{

	   
	   static class FastReader 
	    { 
	        BufferedReader br; 
	        StringTokenizer st; 
	       
	  
	        public FastReader() 
	        { 
	            br = new BufferedReader(new
	                     InputStreamReader(System.in)); 
	        } 
	  
	        String next() 
	        { 
	            while (st == null || !st.hasMoreTokens()) 
	            { 
	                try
	                { 
	                    st = new StringTokenizer(br.readLine()); 
	                } 
	                catch (IOException  e) 
	                { 
	                    e.printStackTrace(); 
	                } 
	            } 
	            return st.nextToken();
	            
	        } 
	  
	        int nextInt() 
	        { 
	            return Integer.parseInt(next()); 
	        } 
	  
	        long nextLong() 
	        { 
	            return Long.parseLong(next()); 
	        } 
	  
	        double nextDouble() 
	        { 
	            return Double.parseDouble(next()); 
	        } 
	  
	        String nextLine() 
	        { 
	            String str = ""; 
	            try
	            { 
	                str = br.readLine(); 
	            } 
	            catch (IOException e) 
	            { 
	                e.printStackTrace(); 
	            } 
	            return str; 
	        }

			public Character charAt(int i) {
				// TODO Auto-generated method stub
				return null;
			}

			public BigInteger nextBigInteger() {
				// TODO Auto-generated method stub
				return null;
			} 
	    } 

  

	  // Complete the hurdleRace function below.
//    public static void solve() {
//    	FastReader s=new FastReader();
	    int binarySearch(int arr[], int l, int r, int x) 
	    { 
	        if (r >= l) { 
	            int mid = l + (r - l) / 2; 
	  
	            // If the element is present at the 
	            // middle itself 
	            if (arr[mid] == x) 
	                return mid; 
	  
	            // If element is smaller than mid, then 
	            // it can only be present in left subarray 
	            if (arr[mid] > x) 
	                return binarySearch(arr, l, mid - 1, x); 
	  
	            // Else the element can only be present 
	            // in right subarray 
	            return binarySearch(arr, mid + 1, r, x); 
	        } 
	  
	        // We reach here when element is not present 
	        // in array 
	        return -1; 
	    } 
	  
	    static // Driver method to test above 

	// Java implementation of iterative Binary Search 

	    // Returns index of x if it is present in arr[], 
	    // else return -1 
	    int lowerBound(int[] a,int n,int key){
	        int s =0,e=n-1;
	        int ans = -1;

	        while(s<=e){
	            int mid = (s+e)/2;

	            if(a[mid]==key){
	                ans = mid;
	                e = mid - 1;
	            }
	            else if(a[mid]>key){
	                e = mid - 1;
	            }
	            else{
	                s = mid + 1;
	            }
	        }

	        return ans;
	    }
	    public static double log2(long N) 
	    { 
	  
	        // calculate log2 N indirectly 
	        // using log() method 
	        double result = (double)(Math.log(N) / Math.log(2)); 
	  
	        return result; 
	    }
    public static void main(String[] args) throws IOException {
    	FastReader s=new FastReader();
    	int t = s.nextInt();
    	while(t-->0) {
    		int n =  s.nextInt();
    		long[] ar = new long[n];
    		for(int i=0;i<n;i++) {
    			ar[i]= s.nextLong();
    		}
    		long max=0;
    		for(int i=1;i<n;i++) {
    			if(ar[i]<ar[i-1]) {
    				max = Math.max(max, ar[i-1]-ar[i]);
    				ar[i]=ar[i-1];
    			}
    		}
    		double p = (double)log2(max+1);
    		System.out.println((int)Math.ceil((double)p));
      	}
   			
    }	
}





        
    

      
 