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.*;

public class PoweredAddition {
    public static int solve(int N, int[] A) {
        int[] max = new int[N], min = new int[N];
        max[0] = A[0];
        min[N - 1] = A[N - 1];
        for (int i = 1; i < N; ++i) {
            max[i] = Math.max(A[i], max[i - 1]);
            min[N - i - 1] = Math.min(A[N - i - 1], min[N - i]);
        }
        int maxdiff = 0;
        for (int i = 1; i < N; ++i) {
            maxdiff = Math.max(max[i] - min[i], maxdiff);
        }
//        System.out.println("maxdiff " + maxdiff);
        if (maxdiff > 0) {
            return (int) (Math.floor(Math.log(maxdiff) / Math.log(2)) + 1);
        }

        return 0;
    }

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int T = input.nextInt();
        for (int ti = 0; ti < T; ++ti) {
            int N = input.nextInt();
            int[] A = new int[N];
            for (int ai = 0; ai < N; ++ai) {
                A[ai] = input.nextInt();
            }
            System.out.println(solve(N, A));
        }
    }
}
