B.MakeItIncreasing

time limit per test

2 seconds

memory limit per test

256 megabytes

Given nn integers a1,a2,…,ana1,a2,…,an. You can perform the following operation on them:

select any element aiai (1≤i≤n1≤i≤n) and divide it by 22 (round down). In other words, you can replace any selected element aiai with the value ⌊ai2⌋⌊ai2⌋ (where ⌊x⌋⌊x⌋ is – round down the real number xx).

Output the minimum number of operations that must be done for a sequence of integers to become strictly increasing (that is, for the condition a1<a2<⋯<ana1<a2<⋯<an to be satisfied). Or determine that it is impossible to obtain such a sequence. Note that elements cannot be swapped. The only possible operation is described above.

For example, let n=3n=3 and a sequence of numbers [3,6,5][3,6,5] be given. Then it is enough to perform two operations on it:

Write the number ⌊62⌋=3⌊62⌋=3 instead of the number a2=6a2=6 and get the sequence [3,3,5][3,3,5];Then replace a1=3a1=3 with ⌊32⌋=1⌊32⌋=1 and get the sequence [1,3,5][1,3,5].

The resulting sequence is strictly increasing because 1<3<51<3<5.

Input

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

The descriptions of the test cases follow.

The first line of each test case contains a single integer nn (1≤n≤301≤n≤30).

The second line of each test case contains exactly nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤2⋅1090≤ai≤2⋅109).

Output

For each test case, print a single number on a separate line — the minimum number of operations to perform on the sequence to make it strictly increasing. If a strictly increasing sequence cannot be obtained, print "-1".

Example

Input

Copy

7

3

3 6 5

4

5 3 2 1

5

1 2 3 4 5

1

1000000000

4

2 8 7 5

5

8 26 5 21 10

2

5 14

Output

Copy

2 -1 0 0 4 11 0

Note

The first test case is analyzed in the statement.

In the second test case, it is impossible to obtain a strictly increasing sequence.

In the third test case, the sequence is already strictly increasing.

解题说明：此题是一道模拟题，按照题目意思，将数列中的数字除以2后保证数列递增。可以从数列队尾往前遍历计算，确保前面的数字小于后面的数字，直到数字为0（如果不是第一个数字，则此时将输出-1）或者遍历到第一个数字为止。

#include<iostream> using namespace std; int main() { int t = 0; cin >> t; be: while (t--) { int n = 0, a[32] = { 0 }, s = 0; cin >> n; for (int i = 0; i < n; i++) { cin >> a[i]; } if (n == 1) { cout << "0\n"; } else { for (int i = n - 2; i >= 0; i--) { if (a[i] == 0 && i != 0 || a[i + 1] == 0) { cout << "-1\n"; goto be; } while (a[i] >= a[i + 1]) { a[i] /= 2; s++; } } cout << s << "\n"; } } return 0; }