C - Powered Addition

C - 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 i1,i2,…,ik which are between 1 and n inclusive, and add 2x−1 to each corresponding position of a. Formally, aij:=aij+2x−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 a1≤a2≤…≤an.

You have to answer t independent test cases.

Input
The first line contains a single integer t (1≤t≤104) — the number of test cases.

The first line of each test case contains single integer n (1≤n≤105) — 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 105.

The second line of each test case contains n integers a1,a2,…,an (−109≤ai≤109).

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].

思路：找到当前位置和前面数的最大差值，把每一个位置的add合并可以得到形如10010的二进制数即差值，假设二进制数的长度为ans，那么它可以表示0~pow（2，ans）-1的数，所以我们只需要求出最大的一个差值需要多少位二进制表示出来即可

#include<iostream>
#include<cmath>
using namespace std;
const int N=1e5+10;
int a[N],qz[N],mx;
int main()
{int t;cin>>t;while(t--){int n;cin>>n;mx=0;for(int i=1;i<=n;i++) scanf("%d",a+i);qz[1]=a[1];for(int i=2;i<=n;i++){qz[i]=max(qz[i-1],a[i]);mx=max(mx,qz[i]-a[i]);}long long ans=0;while(pow(2,ans)-1<mx) ans++;cout<<ans<<endl;}
}

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