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>
#pragma GCC optimize("Ofast")
using namespace std;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline void dump(T& v) {
  for (auto i = v.begin(); i != v.end(); i++) {
    cout << *i << ((i == --v.end()) ? '\n' : ' ');
  }
}
inline string getline() {
  string s;
  getline(cin, s);
  return s;
}
inline void yn(const bool b) {
  b ? cout << ("yes") << '\n' : cout << ("no") << '\n';
}
inline void Yn(const bool b) {
  b ? cout << ("Yes") << '\n' : cout << ("No") << '\n';
}
inline void YN(const bool b) {
  b ? cout << ("YES") << '\n' : cout << ("NO") << '\n';
}
struct io {
  io() {
    ios::sync_with_stdio(false);
    cin.tie(0);
  }
};
const int INF = INT_MAX;
const long long int LLINF = 1LL << 60;
const long long int MOD = 1000000007;
const double EPS = 1e-9;
long long int mod_pow(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x;
    x = x * x;
    n >>= 1;
  }
  return res;
}
int main() {
  int Q;
  cin >> Q;
  while (Q--) {
    long long int n;
    cin >> n;
    vector<long long int> v(n), M(n);
    for (int i = 0; i < (n); ++i) cin >> v[i];
    M[0] = v[0];
    for (int i = 0; i < (n - 1); ++i) M[i + 1] = max(M[i], v[i + 1]);
    long long int diff = 0;
    for (int i = 0; i < (n); ++i) chmax(diff, M[i] - v[i]);
    cout << (ceil(log2(diff + 1))) << '\n';
  }
}
