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 OO = 1e9;
const long long INF = 1e18;
const int irand(int lo, int hi) {
  return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
const long long lrand(long long lo, long long hi) {
  return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
template <typename T>
T getnum() {
  int sign = 1;
  T ret = 0;
  char c;
  do {
    c = getchar();
  } while (c == ' ' || c == '\n');
  if (c == '-')
    sign = -1;
  else
    ret = c - '0';
  while (1) {
    c = getchar();
    if (c < '0' || c > '9') break;
    ret = 10 * ret + c - '0';
  }
  return sign * ret;
}
inline void ini(int& x) { x = getnum<int>(); }
inline void scani(int& x) { scanf("%d", &x); }
const int N = 1e5 + 5;
int t;
int n;
long long arr[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(10);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i += 1) cin >> arr[i];
    long long maxdif = 0;
    long long prev = arr[1];
    for (int i = 2; i <= n; i += 1) {
      maxdif = max(maxdif, prev - arr[i]);
      prev = max(prev, arr[i]);
    }
    int ans = 0;
    long long dif = 0;
    while (1) {
      if (dif >= maxdif) break;
      ans++;
      dif = dif * 2 + 1;
    }
    cout << ans << '\n';
  }
  return 0;
}
