Powered Addition

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 $i_{1}, i_{2}, \ldots, i_{k}$ 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 $j = 1, 2, \ldots, k$. 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 $a_{1} \le a_{2} \le \ldots \le a_{n}$.

You have to answer $t$ independent test cases.Input

The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases.

The first line of each test case contains single integer $n$ ($1 \le n \le 10^{5}$) — 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 $a_{1}, a_{2}, \ldots, a_{n}$ ($-10^{9} \le a_{i} \le 10^{9}$).Output

For each test case, print the minimum number of seconds in which you can make $a$ nondecreasing.Exampleinput

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 $[0, 0]$.

Solution:

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
int val = 0;
int mx = (int) -2e9;
for (int i = 0; i < n; i++) {
      int a;
      cin >> a;
mx = max(mx, a);
val = max(val, mx - a);
}
long long res = 0;
while ((1LL << res) - 1 < val) {
      ++res;
    }
    cout << res << '\n';
  }
  return 0;
}