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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using lld = long double;
using l64 = int64_t;
const ll oo = 0x3f3f3f3f;
const ll MOD = 1000000007;
const ll N = 2e5 + 10;
ll ar[N];
ll req[N];
void solve() {
  ll n;
  cin >> n;
  for (ll i = 0; i < n; i++) cin >> ar[i];
  ll cur_max = -1e15, max_req = -1e15;
  for (ll i = 0; i < n; i++) {
    cur_max = max(cur_max, ar[i]);
    req[i] = cur_max - ar[i];
    req[i] = max(0LL, req[i]);
    max_req = max(max_req, req[i]);
  }
  if (max_req == 0) {
    cout << 0 << '\n';
    return;
  }
  ll x = 0;
  ll ans = 1;
  for (ll i = 0; i < 40; i++) {
    x += (1 << i);
    ans = i;
    if (x >= max_req) break;
  }
  cout << 1 + ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ll tc;
  cin >> tc;
  while (tc--) solve();
  return 0;
}
