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;
int getint(string s) {
  int x = stoi(s);
  return x;
}
string getstring(int x) {
  string lol = to_string(x);
  return lol;
}
bool ok(int x, int gg) { return (x <= gg - 1 and x >= 0); }
int X[4] = {1, 0, -1, 0};
int Y[4] = {0, 1, 0, -1};
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int exponentMod(long long int A, long long int B, long long int C) {
  if (A == 0) return 0;
  if (B == 0) return 1;
  long long int y;
  if (B % 2 == 0) {
    y = exponentMod(A, B / 2, C);
    y = (y * y) % C;
  } else {
    y = A % C;
    y = (y * exponentMod(A, B - 1, C) % C) % C;
  }
  return (long long int)((y + C) % C);
}
bool ispal(string x) {
  for (int i = 0, j = x.size() - 1; i < x.size(); i++, j--) {
    if (i > j) break;
    if (x[j] != x[i]) return false;
  }
  return true;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> v;
    long long int gg[n + 2];
    memset(gg, 0, sizeof gg);
    for (long long int j = 0; j < n; j++) {
      long long int x;
      cin >> x;
      v.push_back(x);
    }
    long long int maxi = 0;
    for (long long int j = 1; j < v.size(); j++) {
      if (v[j] < v[j - 1]) {
        long long int aux = v[j - 1] - v[j];
        for (long long int k = 31; k >= 0; k--) {
          if ((aux & (1 << k)) != 0) {
            maxi = max(maxi, k + 1);
            break;
          }
        }
        v[j] += aux;
      }
    }
    cout << maxi << endl;
  }
}
