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 LIM = 1e5 + 5, MOD = 1e18 + 7;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fastpowMOD(long long a, long long p, long long MOD) {
  if (p == 0) return 1;
  long long z = fastpowMOD(a, p / 2, MOD);
  z = (z * z) % MOD;
  if (p % 2) z = (z * a) % MOD;
  return z;
}
bool seive[100005];
long long dp[10005][105];
void SieveOfEratosthenes(long long n) {
  memset(seive, true, sizeof(seive));
  for (long long p = 2; p * p <= n; p++)
    if (seive[p] == true)
      for (long long i = p * p; i <= n; i += p) seive[i] = false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, i;
  cin >> t;
  for (long long k1 = 1; k1 <= t; k1++) {
    long long n;
    cin >> n;
    long long a[n + 5];
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long max1 = a[0];
    long long max2 = 0;
    for (i = 1; i < n; i++) {
      if (a[i] >= max1) {
        max1 = max(max1, a[i]);
      } else {
        max2 = max(max2, max1 - a[i]);
      }
    }
    long long k = 0;
    long long res = -1;
    while (max2 > 0) {
      if (max2 % 2 == 1) {
        res = max(res, k);
      }
      max2 /= 2;
      k++;
    }
    cout << res + 1;
    cout << endl;
  }
}
