Codeforces Round #633 (Div. 2)C Powered Addition (贪心,二进制)

Powered Addition

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You have an array aa of length nn. For every positive integer xx you are going to perform the following operation during the xx-th second:

• Select some distinct indices i1,i2,…,iki1,i2,…,ik which are between 11 and nn inclusive, and add 2x−12x−1 to each corresponding position of aa. Formally, aij:=aij+2x−1aij:=aij+2x−1 for j=1,2,…,kj=1,2,…,k. Note that you are allowed to not select any indices at all.

You have to make aa nondecreasing as fast as possible. Find the smallest number TT such that you can make the array nondecreasing after at most TT seconds.

Array aa is nondecreasing if and only if a1≤a2≤…≤ana1≤a2≤…≤an.

You have to answer tt independent test cases.

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The first line of each test case contains single integer nn (1≤n≤1051≤n≤105) — the length of array aa. It is guaranteed that the sum of values of nn over all test cases in the input does not exceed 105105.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109−109≤ai≤109).

Output

For each test case, print the minimum number of seconds in which you can make aa 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 3,43,4 at the 11-st second and 44 at the 22-nd second, then aa will become [1,7,7,8][1,7,7,8]. There are some other possible ways to make aa nondecreasing in 22 seconds, but you can't do it faster.

In the second test case, aa is already nondecreasing, so answer is 00.

In the third test case, if you do nothing at first 22 seconds and select index 22 at the 33-rd second, aa will become [0,0][0,0].

思路：找每一个非递增子序列的最大峰值差，比如是1 2 3 4 5 4 1 5 6 4，有两个峰一个是 5 4 1 还有一个是 6 4，取max(5-4,5-1,6-4)就好，然后模拟求出答案~

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rg register ll
#define inf 2147483647
#define lb(x) (x&(-x))
ll sz[200005],n;
template <typename T> inline void read(T& x)
{
    x=0;char ch=getchar();ll f=1;
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}x*=f;
}
inline ll query(ll x){ll res=0;while(x){res+=sz[x];x-=lb(x);}return res;}
inline void add(ll x,ll val){while(x<=n){sz[x]+=val;x+=lb(x);}}//第x个加上val
ll t,a[100005];
int main() 
{
	cin>>t;
	for(int i=1;i<=t;i++)
	{
		ll x;
		cin>>x;
		for(int i=1;i<=x;i++)cin>>a[i];
		ll maxx=-1,ans=0;
		for(int i=1;i<x;i++)
		{
			if(a[i]>a[i+1])
			{
				int j;
				for(j=i+1;a[i]>=a[j]&&j<=x;j++)
				{
					maxx=max(a[i]-a[j],maxx);
				}
				i=j-1;
			}
		}
		//cout<<maxx<<endl;
		if(maxx<=0)
		{
			cout<<0<<endl;
			continue;
		}
		int k=1;
		while(maxx>0)
		{
			maxx-=k;
			ans++;
			k*=2;
		}
		cout<<ans<<endl;
		
	}
    return 0;
    
}