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;
void func(long long s, long long e, vector<int> &ans, long long n,
          vector<long long> arr) {
  if (e == 0) return;
  long long i;
  long long max = INT_MIN;
  long long maxIdx = -1;
  for (i = s; i < e; i++) {
    if (max < arr[i]) {
      max = arr[i];
      maxIdx = i;
    }
  }
  long long tempMIN = max;
  for (i = maxIdx + 1; i < e; i++) {
    if (tempMIN > arr[i]) tempMIN = arr[i];
  }
  long long diff = max - tempMIN;
  int secReq = 0;
  if (diff != 0) secReq = (int)log2(diff) + 1;
  ans.push_back(secReq);
  func(0, maxIdx, ans, n, arr);
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long i, val;
    vector<long long> arr;
    vector<long long> copyIT;
    long long n;
    cin >> n;
    vector<int> ans;
    for (i = 0; i < n; i++) {
      cin >> val;
      arr.push_back(val);
      copyIT.push_back(val);
    }
    sort(copyIT.begin(), copyIT.end());
    if (arr == copyIT)
      cout << 0 << "\n";
    else {
      func(0, n, ans, n, arr);
      sort(ans.begin(), ans.end(), greater<int>());
      cout << ans[0] << "\n";
    }
  }
  return 0;
}
