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 solve();
void solveA();
void solveB();
void solveC();
void solveD();
void solveE();
void solveF();
void solveG();
void solveTests() {
  int nTests;
  cin >> nTests;
  for (int i = 0; i < nTests; i++) {
    solve();
  }
}
template <typename T>
bool is_even(T n) {
  return (n % 2 == 0);
}
template <typename T>
bool is_odd(T n) {
  return (n % 2 == 1);
}
template <typename T>
T get_max_in_vector(vector<T> &v) {
  T mx = v[0];
  for (T tmp : v) mx = max(mx, tmp);
  return mx;
}
template <typename T>
T get_min_in_vector(vector<T> &v) {
  T mn = v[0];
  for (T tmp : v) mn = min(mn, tmp);
  return mn;
}
int main() {
  solveTests();
  return 0;
}
void solve() { solveC(); }
long long get_ones_count(vector<long long> &v, long long tmp) {
  long long count = 0;
  for (int i = 0; i < v.size(); i++) {
    if ((v[i] ^ tmp) > 0) count++;
  }
  return count;
}
void informatics_77() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (auto &x : v) {
    cin >> x;
  };
  long long tmp = 1;
  long long sum = 0;
  for (int i = 0; i < 16; i++) {
    long long ones_count = get_ones_count(v, tmp);
    sum += (tmp * ones_count * (n - ones_count));
    tmp *= 2;
  }
  cout << sum << endl;
}
void informatics_282() {}
void solveA() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (auto &x : v) {
    cin >> x;
  };
  sort(v.begin(), v.end());
  long long l = n / 2 - 1;
  long long r = n / 2;
  bool l_turn = false;
  while (l >= 0 || r < n) {
    if (l_turn) {
      cout << v[l--] << " ";
    } else
      cout << v[r++] << " ";
    l_turn = !l_turn;
  }
}
void solveB() {}
void solveC() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (auto &x : v) {
    cin >> x;
  };
  long long max_before = v[0];
  long long max_def = 0;
  for (int i = 0; i < n; i++) {
    max_def = max(max_def, max_before - v[i]);
    max_before = max(max_before, v[i]);
  }
  if (max_def == 0) {
    cout << 0 << endl;
    return;
  }
  long long sum = 0;
  long long tmp = 1;
  long long pow = 0;
  while (sum < max_def) {
    sum += tmp;
    pow++;
    tmp *= 2;
  }
  cout << pow << endl;
}
void solveD() {}
void solveE() {}
