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>
const double PI = 3.141592653589793238463;
using namespace std;
inline long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
inline long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += 1000000007;
  return a;
}
inline long long mul(long long a, long long b) {
  return (long long)((long long)a * b % 1000000007);
}
long long fast_power(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power % 2 == 1) {
      result = (result * base) % 1000000007;
    }
    base = (base * base) % 1000000007;
    power = power / 2;
  }
  return result;
}
long long min(long long a, long long b) { return a > b ? b : a; }
long long max(long long a, long long b) { return a > b ? a : b; }
void SieveOfEratosthenes(long long n, unordered_set<long long> &s) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long p = 2; p <= n; p++)
    if (prime[p]) s.insert(p);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long mx;
    long long tmp;
    long long diff = 0;
    for (long long i = 0; i < n; i++) {
      cin >> tmp;
      if (i == 0) {
        mx = tmp;
      }
      mx = max(mx, tmp);
      diff = max(diff, mx - tmp);
    }
    long long ans = 0;
    if (diff == 0) {
      cout << 0 << "\n";
    } else {
      long long val = 1;
      while (diff > 0) {
        diff -= val;
        val *= 2;
        ans += 1;
      }
      cout << ans << "\n";
    }
  }
}
