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 int mpow(long long int base, long long int exp);
void ipgraph(long long int m);
void dfs(long long int u, long long int par);
const long long int mod = 1000000007;
const long long int N = 3e5, M = N;
vector<long long int> g[N];
struct data {
  bool pos;
  bool neg;
};
long long int fib(long long int n) {
  long long int a = 1, b = 1, c, i;
  if (n == 0) return a;
  for (i = 2; i <= n; i++) {
    c = a + b;
    a = b;
    b = c;
  }
  return b;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t, i, j, k, p, q, r, x, y, u, v, n, m;
  cin >> t;
  while (t--) {
    cin >> n;
    long long int a[n + 1];
    for (long long int i = 1; i <= n; i++) cin >> a[i];
    long long int ans = 0;
    long long int tillmax = a[1];
    for (long long int i = 1; i <= n; i++) {
      tillmax = max(tillmax, a[i]);
      ans = max(ans, tillmax - a[i]);
    }
    long long int fans = ceil((double)log2(ans + 1));
    cout << fans << endl;
  }
}
long long int mpow(long long int base, long long int exp) {
  base %= mod;
  long long int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long int)result * base) % mod;
    base = ((long long int)base * base) % mod;
    exp >>= 1;
  }
  return result;
}
void ipgraph(long long int n, long long int m) {
  long long int i, u, v;
  while (m--) {
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
}
void dfs(long long int u, long long int par) {
  for (long long int v : g[u]) {
    if (v == par) continue;
    dfs(v, u);
  }
}
