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 power(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a);
    a = (a * a);
    b >>= 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, i;
    cin >> n;
    long long a[n + 1];
    long long diff;
    for (i = 1; i <= n; i++) cin >> a[i];
    long long cnt = 0;
    long long tt = 1, ot = 0;
    long long maxx = a[1], maxd = 0;
    for (i = 2; i <= n; i++) {
      if (a[i] < maxx) {
        diff = maxx - a[i];
        cnt++;
        maxd = max(maxd, diff);
      } else
        maxx = max(maxx, a[i]);
    }
    if (!cnt)
      cout << 0 << endl;
    else {
      vector<long long> v;
      long long pr = 1, prev = 1, cnt = 1, cuml = 1;
      while (cuml < maxd) {
        pr <<= 1;
        cuml = pr + prev;
        cnt++;
        prev = cuml;
      }
      cout << cnt << endl;
    }
  }
}
