Codeforces Round 961 (Div. 2) C. Squaring

Ikrpprpp找到了一个由整数组成的数组 a a a 。他喜欢正义，所以他想要公平，也就是说，让它不递减。要做到这一点，他可以对数组的索引 1 ≤ i ≤ n 1 \le i \le n 1≤i≤n 执行一个公正的行为，它将用 a i 2 a_i ^ 2 ai2 替换 a i a_i ai (位置 i i i 的元素及其平方)。例如，如果 a = [ 2 , 4 , 3 , 3 , 5 , 3 ] a = [2,4,3,3,5,3] a=[2,4,3,3,5,3] 和ikrpprpp选择对 i = 4 i = 4 i=4 执行正义行为，则 a a a 变为 [ 2 , 4 , 3 , 9 , 5 , 3 ] [2,4,3,9,5,3] [2,4,3,9,5,3] 。

使数组不递减所需的最少正义行为是多少?

Input

First line contains an integer t t t ( 1 ≤ t ≤ 1000 1 \le t \le 1000 1≤t≤1000) --- the number of test cases. It is followed by the description of test cases.

For each test case, the first line contains an integer n n n --- size of the array a a a. The second line contains n n n ( 1 ≤ n ≤ 2 ⋅ 1 0 5 1 \le n \le 2 \cdot 10 ^5 1≤n≤2⋅105) integers a 1 , a 2 , ... , a n a_1, a_2,\ldots, a_n a1,a2,...,an ( 1 ≤ a i ≤ 1 0 6 1 \le a_i \le 10 ^ 6 1≤ai≤106).

The sum of n n n over all test cases does not exceed 2 ⋅ 10 5 2 \cdot {10}^5 2⋅105.

Output

For each testcase, print an integer --- minimum number of acts of justice required to make the array a a a non-decreasing. If it is impossible to do that, print − 1 -1 −1.

Example

cpp 复制代码

7
3
1 2 3
2
3 2
3
3 1 5
4
1 1 2 3
3
4 3 2
9
16 2 4 2 256 2 4 2 8
11
10010 10009 10008 10007 10006 10005 10004 10003 10002 10001 10000

output

cpp 复制代码

Note

In the first test case, there's no need to perform acts of justice. The array is fair on its own!

In the third test case, it can be proven that the array cannot become non-decreasing.

In the fifth test case, ikrpprppp can perform an act of justice on index 3, then an act of justice on index 2, and finally yet another act of justice on index 3. After that, a a a will become [ 4 , 9 , 16 ] [4, 9, 16] [4,9,16].

cpp 复制代码

#include<bits/stdc++.h>  
using namespace std;  

typedef pair<int,int> PII;
typedef long long LL;

inline void solve()
{
	int n;cin>>n;
	vector<LL> a(n);
	for(int i=0;i<n;i++) cin>>a[i];
	
	int cnt=0;
	LL res=0;
	for(int i=1;i<n;i++)
	{
		if(a[i]==1 && a[i-1]>1) 
		{
			cout<<"-1\n";
			return ;
		}
		LL t=a[i];
		while(a[i-1]>t) 
		{
			t=t*t;
			cnt++;
		}
		while(cnt>0 && a[i-1]*a[i-1]<=t)
		{
			t=sqrtl(t);
			cnt--;
		}
		res+=cnt;
	}
	cout<<res<<"\n";
}

signed main() 
{  
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int T=1;cin>>T;
	while(T--) solve();
    
    return 0;
}