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;
using ll = long long;
using db = double;
using ld = long double;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = vector<vector<T>>;
constexpr ll mod = 1000000007;
constexpr ll INF = 1000000099;
constexpr ll LINF = (ll)(1e18 + 99);
const vector<ll> dx = {-1, 1, 0, 0}, dy = {0, 0, -1, 1};
template <typename T, typename U>
inline bool chmin(T& t, const U& u) {
  if (t > u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T, typename U>
inline bool chmax(T& t, const U& u) {
  if (t < u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T, typename Y>
inline T mpow(T a, Y n) {
  T res = 1;
  for (; n; n >>= 1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  int qq;
  cin >> qq;
  while (qq--) {
    ll n, mx, dif = 0;
    cin >> n;
    for (ll i = 0; i < (ll)(n); ++i) {
      ll a;
      cin >> a;
      if (i) {
        chmax(dif, mx - a);
        chmax(mx, a);
      } else {
        mx = a;
      }
    }
    for (ll i = 0; i < (ll)(33); ++i) {
      if (dif <= 0) {
        cout << i << '\n';
        break;
      } else {
        dif -= 1ll << i;
      }
    }
  }
}
