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.*;
import java.util.stream.Collectors;

public class Solution {

    private static final String NO = "NO";
    private static final String YES = "YES";
    public static BufferedWriter bw;
    public static FR sc;
    private static final int LIMIT = (int) Math.pow(10, 6);
    private static long[] dp;

    public static void main(String[] args) throws Exception {
        int size = 40;
        dp = new long[size];
        dp[1] = 1;
        for (int i = 2; i < size; i++) {
            dp[i] = dp[i - 1] * 2;
        }

        FR sc = new FR();
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
        int t = sc.i();
        while (t > 0) {
            t--;
            int n = sc.i();
            long[] a = sc.la(n);
            int ans = solve(n, a);
            bw.write(ans + "\n");
        }
        bw.close();
    }

    private static int solve(int n, long[] a) {
        int ans = 0;
        for (int i = 1; i < n; i++) {
            if (a[i] < a[i - 1]) {
                long diff = Math.abs(a[i] - a[i - 1]);
                for (int j = 1; j < 40; j++) {
                    if (dp[j] >= diff) {
                        int temp = j;
                        if (dp[j] != diff) temp--;
                        a[i] = a[i - 1];
                        if (temp > ans) {
                            ans = temp;
                        }
                        break;
                    }
                }
            }

        }
        return ans;
    }


}

class FR {
    BufferedReader br;
    StringTokenizer st;

    public FR() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    public String s() {
        if (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {

            }
        }
        return st.nextToken();
    }

    public int i() {
        return Integer.parseInt(s());
    }

    public int[] ia(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = i();
        }
        return a;
    }

    public long[] la(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = l();
        }
        return a;
    }

    public String[] sa(int n) {
        String[] a = new String[n];
        for (int i = 0; i < n; i++) {
            a[i] = s();
        }
        return a;
    }

    public long l() {
        return Long.parseLong(s());
    }
}