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 int INF = 2147483647;
const int INF2 = 0x3f3f3f3f;
const long long INF64 = 1e18;
const double INFD = 1e30;
const double EPS = 1e-9;
const double PI = acos(-1);
const int MOD = 998244353;
template <typename T>
inline T read() {
  T X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
const int MAXN = 100005;
int n, m, k;
int dr[4] = {1, 0, -1, 0};
int dc[4] = {0, 1, 0, -1};
int dr2[8] = {1, 1, 1, -1, -1, -1, 0, 0};
int dc2[8] = {1, 0, -1, 1, 0, -1, 1, -1};
int CASE = 1;
int arr[MAXN];
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
  }
  int curmax = arr[1];
  int num = 0;
  for (int i = 2; i <= n; i++) {
    if (arr[i] < curmax) {
      num = max(num, curmax - arr[i]);
    }
    curmax = max(curmax, arr[i]);
  }
  if (!num) {
    printf("0\n");
  } else {
    long long ans = 0;
    for (long long i = 1; i <= num; i *= 2) {
      ans++;
    }
    printf("%lld\n", ans);
  }
}
int main() {
  int T = read<int>();
  while (T--) {
    solve();
  }
  return 0;
}
