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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long power_mod(long long x, long long y, long long m) {
  long long ans = 1;
  x %= m;
  while (y) {
    if (y & 1) {
      ans = (ans * x) % m;
    }
    y >>= 1;
    x = (x * x) % m;
  }
  return ans;
}
long long power(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) {
      ans *= x;
    }
    y >>= 1;
    x = x * x;
  }
  return ans;
}
long long mod_inv(long long x, long long m) { return power_mod(x, m - 2, m); }
int randomGenerator(int limit) {
  random_device rd;
  mt19937 gen(rd());
  uniform_int_distribution<> dis(1, limit);
  return dis(gen);
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
bool comp(const pair<long long, pair<long long, long long>> &a,
          const pair<long long, pair<long long, long long>> &b) {
  if (a.first == b.first) {
    return (a.second < b.second);
  }
  return (a.first < b.first);
}
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2> &p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool isPalindrome(string s) {
  bool flag = true;
  for (int i = 0; i < s.length() / 2; i++) {
    if (s[i] != s[s.length() - i - 1]) {
      flag = false;
      break;
    }
  }
  return flag;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < (n); i++) {
    cin >> v[i];
  }
  long long mn = 1e18;
  long long mx = v[0];
  for (long long i = (1); i <= (n - 1); i++) {
    long long d = (v[i] - mx);
    if (d < 0) {
      mn = min(mn, d);
    } else {
      mx = max(mx, v[i]);
    }
  }
  if (mn == 1e18) {
    mn = 0;
  }
  long long ans = ceil(log2(abs(mn) + 1));
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
