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;
const long long int mod = 1e9 + 7;
const long long int mod1 = 998244353;
void swap(long long int &a, long long int &b) {
  long long int t = a;
  a = b;
  b = t;
}
long long int min(long long int a, long long int b) {
  if (a < b) return a;
  return b;
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int pow(long long int a, long long int b) {
  a %= mod;
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return ans;
}
int n;
int visited[102] = {0};
int x[102], y[102];
void dfs(int i) {
  visited[i] = 1;
  for (int j = 0; j < n; j++) {
    if (x[j] == x[i] || y[j] == y[i]) {
      if (visited[j] == 0) {
        dfs(j);
      }
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    int sorted = 1;
    for (long long int i = 1; i < n; i++) {
      if (a[i] < a[i - 1]) {
        sorted = 0;
        break;
      }
    }
    if (sorted) {
      cout << "0\n";
      continue;
    }
    long long int time = 1;
    long long int lastTotal = 1;
    long long int lastPower = 1;
    for (long long int i = 1; i < n; i++) {
      int cur = 1, pow = 0;
      long long int temp = a[i];
      while (a[i] < a[i - 1]) {
        a[i] += 1 << pow;
        pow++;
      }
      if (pow > lastPower) {
        lastPower = pow;
      }
      pow--;
      int red = 0;
      while (pow >= 0) {
        a[i] -= 1 << pow;
        if (a[i] < a[i - 1]) {
          a[i] += 1 << pow;
        }
        pow--;
      }
    }
    cout << lastPower << endl;
  }
  return 0;
}
