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].
#include <bits/stdc++.h>
using namespace std;
long long ar[31];
int squar() {
  for (int i = 0; i <= 30; i++) ar[i] = pow(2, i) - 1;
  return 0;
}
int upper(long long arr[], int n, int s) {
  int u = s - 1, l = 0;
  int found = -1;
  int mid = (u + l) / 2;
  while (u >= l) {
    if (arr[mid] > n)
      u = mid - 1;
    else if (arr[mid] < n)
      l = mid + 1;
    else if (arr[mid] == n) {
      found = mid;
      break;
    }
    mid = (u + l) / 2;
  }
  if (found == -1) {
    return l;
  } else
    return mid;
}
int solve() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  int max = arr[0], ind = -1, min, ans = 0;
  for (int i = 1; i < n; i++) {
    if (arr[i] >= arr[i - 1]) {
      if (ind != -1) ans = (max - min > ans ? max - min : ans);
      ind = -1;
    } else {
      if (ind == -1) {
        max = (arr[i - 1] > max ? arr[i - 1] : max);
        ind = i - 1;
        min = arr[i];
      } else
        min = arr[i];
    }
  }
  if (ind != -1) ans = (max - min > ans ? max - min : ans);
  int y = upper(ar, ans, 31);
  cout << y << '\n';
  return 0;
}
int main() {
  squar();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t) {
    solve();
    t--;
  }
  return 0;
}
