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;
using ll = long long;
using ull = unsigned long long;
template <typename T, typename A>
int arg_max(std::vector<T, A> const& vec) {
  return static_cast<int>(
      std::distance(vec.begin(), max_element(vec.begin(), vec.end())));
}
template <typename T, typename A>
int arg_min(std::vector<T, A> const& vec) {
  return static_cast<int>(
      std::distance(vec.begin(), min_element(vec.begin(), vec.end())));
}
vector<int> binary2(ll x) {
  vector<int> res;
  while (x > 0) {
    res.push_back(x % 2);
    x >>= 1;
  }
  return res;
}
void primeFactors(vector<int>& v, int n) {
  while (n % 2 == 0) {
    v.push_back(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      v.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) v.push_back(n);
}
vector<int> fct(int a, int b) {
  int it = 0;
  while (b < a) {
    b = pow(2, it + 1) - 1;
    it++;
  }
  return {b, it};
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  ll tt;
  cin >> tt;
  for (ll tti = 0; tti < tt; ++tti) {
    int n;
    cin >> n;
    vector<int> v(n, 0);
    ll res = 0;
    for (ll i = 0; i < n; ++i) {
      cin >> v[i];
    }
    int free = 0;
    int maxdiff = 0;
    int maxi = v[0];
    for (ll i = 0; i < n - 1; ++i) {
      maxdiff = max(maxdiff, maxi - v[i + 1]);
      maxi = max(maxi, v[i + 1]);
    }
    vector<int> r = fct(maxdiff, 0);
    cout << r[1] << endl;
  }
  return 0;
}
