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;
const int N = 1e5 + 5;
int tc;
int n;
long long a[N], b[N];
bool can(int x) {
  for (int i = 1; i <= n; i++) {
    b[i] = a[i];
  }
  long long addz = (1LL << x) - 1;
  b[n] += addz;
  for (int i = n - 1; i >= 1; i--) {
    if (b[i] > b[i + 1]) return false;
    b[i] = min(b[i + 1], b[i] + addz);
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    int l = 0, r = 60, ans = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (can(mid)) {
        r = mid - 1;
        ans = mid;
      } else {
        l = mid + 1;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
