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;
long long readn() {
  long long sum = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = sum * 10 + (ch - '0');
    ch = getchar();
  }
  return sum * f;
}
int n, A[110000], Maxn[110000], Minn[110000], t;
void opk() {
  t = readn();
  while (t--) {
    n = readn();
    for (int i = 1; i <= n; i++) {
      A[i] = readn();
      if (i == 1)
        Maxn[i] = A[i];
      else
        Maxn[i] = max(Maxn[i - 1], A[i]);
    }
    Minn[n] = A[n];
    for (int i = n - 1; i > 0; i--) Minn[i] = min(Minn[i + 1], A[i]);
    int ans = 0;
    for (long long i = 1; i <= n; i++) ans = max(ans, Maxn[i] - Minn[i]);
    int p = 0, ans1 = 0;
    while (ans - p > 0) {
      ans1++;
      p = p * 2 + 1;
    }
    printf("%d\n", ans1);
  }
}
signed main() { opk(); }
