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;
inline int add(int a, int b) {
  a += b;
  return a >= 1000000007 ? a - 1000000007 : a;
}
inline int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + 1000000007 : a;
}
inline int mul(int a, int b) { return (long long int)a * b % 1000000007; }
int Set(int N, int pos) { return N = N | (1 << pos); }
int Reset(int N, int pos) { return N = N & ~(1 << pos); }
bool Cheek(int N, int pos) { return (bool)(N & (1 << pos)); }
long long int ara[100005], mx[100005], mn[100005], pr[60];
void Solve(int t) {
  long long int i, j, k, n, mxdif = 0, ans, x;
  scanf("%lld", &(n));
  mx[0] = -92233720368547758;
  for (__typeof(i) i = 1; i <= n; i++) {
    scanf("%lld", &(ara[i]));
    mx[i] = max(ara[i], mx[i - 1]);
  }
  mn[n + 1] = 92233720368547758;
  for (i = n; i >= 1; i--) {
    mn[i] = min(ara[i], mn[i + 1]);
  }
  for (i = 1; i <= n; i++) {
    if (mx[i] > mn[i]) mxdif = max(mxdif, mx[i] - mn[i]);
  }
  if (mxdif == 0)
    ans = 0;
  else {
    for (i = 1; i <= 40; i++) {
      if (pr[i] >= mxdif) {
        ans = i;
        break;
      }
    }
  }
  cout << ans << endl;
}
int main() {
  int t, T;
  long long int val = 1;
  for (long long int i = 1; i <= 40; i++) {
    val *= 2;
    pr[i] = val - 1;
  }
  scanf("%d", &(T));
  for (__typeof(t) t = 1; t <= T; t++) {
    Solve(t);
  }
  return 0;
}
